Connection between commutative algebra and topology
Sumit Kumar Upadhyay$^1$
Shiv Datt Kumar$^2$ AND RAJA SRIDHARAN$^3$\vspace{.4cm
The main aim of this paper to show how commutative algebra is connected to topology. We give underlying topological idea of some results on completable unimodular rows. %Suppose a unimodular row in a Noetherian ring $A$ is elementary completable. Then this unimodular row gives a continuous map which is homotopic to a constant map. %The main aim of this paper is to give Underlying topological idea of some results on unimodular rows. We also give $I/I^2$ anolouge, where $I$ is a proper ideal of $A$, of some results over uimodular row. In this paper we show how commutative algebra is connected to topology and what is meaning of a global results in local sense.
1506.02356.tex
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\title{Connection between commutative algebra and topology}
\author{ Sumit Kumar Upadhyay$^1$, Shiv Datt Kumar$^2$ AND RAJA SRIDHARAN$^3$\vspace{.4cm}\\
$^{1,2}$DEPARTMENT OF MATHEMATICS\\
MOTILAL NEHRU NATIONAL INSTITUTE OF TECHNOLOGY \\
ALLAHABAD, U. P., INDIA \vspace{.3cm}\\ $^3$School Of Mathematics,\\ Tata Institute of fundamental Research,\\Colaba,  Mumbai, India}
%\thanks{ $^1$upadhyaysumit365@gmail.com, $^2$sdt@mnnit.ac.in}
\thanks {2010 Mathematics Subject classification : 13C10, 16D40 }
\thanks{E-mail: $^1$upadhyaysumit365@gmail.com, $^2$sdt@mnnit.ac.in, $^3$sraja@tifr.math.res.in}.\\
\thanks{ $^*$This work is supported by Council of Scientific and Industrial Research grant.}
\begin{abstract}
The main aim of this paper to show how commutative algebra is connected to topology. We give underlying topological idea of some results on completable
unimodular rows.
%Suppose a unimodular row in a Noetherian ring $A$ is elementary completable. Then this unimodular row gives a continuous map which is homotopic to a constant map.
%The main aim of this paper is to give Underlying topological idea of some results on unimodular rows. We also give $I/I^2$  anolouge, where $I$ is a proper ideal of $A$, of some results over uimodular row. In this paper we show how commutative algebra is connected to topology and what is meaning of a global results in local sense.   
\end{abstract}
\maketitle
\hspace{0.8cm}\textbf{Keywords}:  Unimodular row, Matrix, Homotopy.
\section{Introduction}
Let $k$ be a field and  $\vec{x} = (x_1, x_2 \cdots, x_n)\in k^n$ be a non-zero vector, then $\vec{x}$ can be completed to a basis of $k^{n}$. We wish to have an analogue of the above statement for rings.
Let $A$ be a ring and $\vec{a} \in A^{n}, \vec{a} \ne 0$. Then there is a natural question when can $\vec{a}$ be completed to a basis of $A^{n}$?
 Suppose  $\vec{a} = (a_1, a_2, \cdots, a_n)$ can be completed to a basis  of $A^{n}$. Consider these basis vectors as columns of a matrix $\alpha$, whose first column is 
$(a_1, a_2, \cdots, a_n)$. The fact that these vectors span $A^{n}$ imply that there exists a $n\times n$ invertible matrix $\beta$ such that $\alpha\beta = I_n$.
Conversely if there exists an invertible matrix $\alpha \in M_n(A)$ with  first column $(a_1, a_2, \cdots, a_n)$, then $(a_1, a_2, \cdots, a_n)$ can be completed to a basis of $A^{n}$. 
% Therefore $(f_1, f_2, \cdots, f_n)$ can be completed to a basis of $A^{n}$ if and only if $(f_1, f_2, \cdots, f_n)$ is the first row of a matrix in $GL_n(A)$. 
%Necessary condition for this to be possible as can be seen by expanding the determinant of the matrix along the first row is that the ideal generated by $a_1, a_2, \cdots, a_n$
%should contain a unit of $A$. 
%In otherwords, that the ideal generated by $\langle a_1, a_2, \cdots, a_n\rangle = A$.
%  \begin{definition}
% A row $(f_1, f_2, \cdots, f_n) \in A^{n}$ is said to be unimodular if the ideal $\langle f_1, f_2, \cdots, f_n\rangle = A$. Set of all 
% unimodular rows of length $n$ is denoted by $Um_n(A)$.
% \end{definition}
Since any completable row $(a_1, a_2, \cdots, a_n)$ is unimodular, this leads to the following problem : %, which asks ``Is this necessary condition sufficient?"
%\noindent \textbf{Problem}: 
Suppose $(a_1, a_2, \cdots, a_n) \in A^{n}$ is a unimodular row. Then  can one complete the row $(a_1, a_2, \cdots, a_n)$ to a matrix belonging to $GL_n(A)$? 
  
In general answer  of this question is negative. Surprisingly this is related to topology. 
%We remark that if one can complete $(a_1, a_2, \cdots, a_n)$ to a matrix
%$\alpha$ in $GL_n(A)$ having determinant a unit $u\in A$ and having first row $(a_1, a_2, \cdots, a_n)$, then multiplying the last row of $\alpha$ by $u^{-1}$, it follows that 
%$(a_1, a_2, \cdots, a_n)$ can be completed to a matrix of determinant $1$.
%\begin{definition}
% We define $SL_n(A)$ to be the subgroup of $GL_n(A)$ consisting of matrices whose determinant $1$.
%\end{definition}
Suppose one can find a matrix $\alpha \in GL_n(A)$ having first column $(a_1, a_2, \cdots, a_n)$. Then $e_1 \alpha^t = (a_1, a_2, \cdots, a_n)$, where $e_1 = (1, 0, \cdots, 0)$. The group $GL_n(A)$  acts on $A^{n}$ via matrix multiplication. The row $(a_1, a_2, \cdots, a_n)$ can be completed to a
matrix in $GL_n(A)$ if and only if $(a_1, a_2, \cdots, a_n)$ lies in the  orbit of $(1, 0, \cdots, 0)$ under the $GL_n(A)$ action (a similar statement holds for $SL_n(A)$).
% In fact 
%if \\$v = (a_1, a_2, \cdots, a_n)\in A^{n}, \alpha \in GL_n(A)$ and $v \alpha = (b_1, b_2, \cdots, b_n)$ then ideal \\$\langle b_1, b_2, \cdots, b_n\rangle \subset
%\langle a_1, a_2, \cdots, a_n\rangle$. Since $v = (b_1, b_2, \cdots, b_n) \alpha^{-1}$, we have\\ $\langle a_1, a_2, \cdots, a_n\rangle \subset
%\langle b_1, b_2, \cdots, b_n\rangle$. Hence $\langle a_1, a_2, \cdots, a_n\rangle = \langle b_1, b_2, \cdots, b_n\rangle$.
%Therefore the completability of a unimodular row to a matrices in $GL_n(A)$ is equivalent to asking whether 
% $(1, 0, \cdots, 0)$ lies in the $GL_n(A)$ orbit of that unimodular row. The above discussion also holds for the group $SL_n(A)$.
\begin{example}\label{1}
 Let $(x_1, x_2) \in \mathbb{Z}^{2}$. Then $(x_1, x_2)$ is unimodular if and only if $x_1, x_2$ are relatively prime. In this case there exist $x, y \in \mathbb{Z}$
such that $x_{1}x + x_{2}y = 1$ and the matrix  $\left(\begin{array}{cccc} x_{1}& -y\\ x_2 & x\end{array}\right)$ has determinant $1$.
We can find an explicit completion of $(x_1, x_2)$ using the Euclidean Algorithm in the following manner: Assume for simplicity that 
$x_1, x_2 > 0$ and $x_1 > x_2$.
Then by division algorithm  $x_1 = x_{2}q + r$, where $q$ is the quotient and $r$ is the remainder. Then 
$\left(\begin{array}{cccc} 1& -q\\ 0 & 1\end{array}\right) \left(\begin{array}{cccc} x_{1}\\ x_2 \end{array}\right) = 
\left(\begin{array}{cccc} r\\ x_2 \end{array}\right)$. It follows by iterating the above procedure that we can get a matrix $\alpha$ which is a product of matrices of 
the form $\left(\begin{array}{cccc} 1& q\\ 0 & 1\end{array}\right)$, $\left(\begin{array}{cccc} 1& 0\\ q' & 1\end{array}\right)$ where $q, q' \in \mathbb{Z}$ such that 
$\alpha \left(\begin{array}{cccc} x_1\\ x_2 \end{array}\right) = \left(\begin{array}{cccc} 1\\ 0\end{array}\right)$. Then $\alpha^{-1} \left(\begin{array}{cccc} 1\\ 0 \end{array}\right) = \left(\begin{array}{cccc} x_1\\ x_2\end{array}\right)$ and $\alpha \in SL_2(\mathbb{Z})$.
\end{example}
This example motivates to define $E_n(A)$ (see 2.6.3).
\begin{definition}
 Let $A$ be a commutative ring with identity. Let $e_{ij}(\lambda), i\ne j$ be the $n \times n$ matrix in $SL_n(A)$ which has $1$  as its diagonal entries and $\lambda$
as its $(i, j)^{th}$ entry. Let $E_n(A)$ be the subgroup of $SL_n(A)$ generated by $e_{ij}(\lambda), i\ne j$. We call elements of $E_n(A)$ as elementary matrices.
 \end{definition}
For any Euclidean domain $A$, any unimodular row $(a_1, a_2, \cdots, a_n)\in A^{n}$ can be completed to an elementary matrix, where $n\geq 2$ \textit{i.e.} $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (1,0, \cdots,0)$, where $\stackrel {E_n(A)}\sim$ denotes the induced action of $E_n(A)$ on unimodular rows. 

Note that $E_n(A)\subset SL_n(A) \subset GL_n(A)$, hence $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (1,0, \cdots,0) \Rightarrow 
(a_1, a_2, \cdots, a_n) \stackrel {SL_n(A)}\sim (1,0, \cdots,0) \Rightarrow (a_1, a_2, \cdots, a_n) \stackrel {GL_n(A)}\sim (1,0, \cdots,0)$. Since $A = k[X]$ (where $k$ is a field) is a Euclidean domain, $(a_1, a_2, \cdots, a_n)\stackrel {E_n(A)}\sim (1,0, \cdots,0)$ for any unimodular row $(a_1, a_2$, $\cdots, a_n)$ $\in A^{n}$.
\begin{proposition}\label{2}
 Let $A$ be a ring and $(a_1, a_2$, $\cdots, a_n)$ $\in A^{n}$ be a unimodular row. Then $(a_1+\lambda X a_{2}, a_2,\cdots, a_n) \in {A[X]}^{n}$ is a unimodular row over $A[X]$, for every $\lambda \in A$.
\end{proposition}
\begin{proof}
 Since $(a_1, a_2, \cdots, a_n) \in A^{n}$ is a unimodular, there exists $(b_1, b_2, \cdots, b_n) \in A^{n}$ such that $\sum_{i = 1}^{n} a_{i}b_{i} = 1$. Take 
$c_1 = b_1, c_2 = b_2 - b_{1}\lambda X, c_3 = b_3, \cdots, c_n = b_n$, thus we have $c_{1}(a_1+\lambda Xa_{2}) + \sum_{i = 2}^{n} a_{i}c_{i} = 1$. Therefore 
$(a_1+\lambda X a_{2}, a_2,\cdots, a_n)$ is a unimodular row over $A[X]$.
\end{proof}
From Proposition \ref{2}, we can prove that for any matrix $\sigma = \prod_{i = 1}^{r} E_{ij}(\lambda) \\ \in E_n(A)$, $(a_1, a_2, \cdots, a_n)\sigma(X) \in {A[X]}^{n}$ is a unimodular over $A[X]$, where $\sigma(X) = \prod_{i = 1}^{r} E_{ij}(\lambda X)$.
\section{Topological fact}
\begin{lemma}\label{3}
 Let $T$ be a topological space. Then any map $f : T \longrightarrow T$ is homotopic to itself.
\end{lemma}
\begin{lemma}[\cite{JR}, Exercise 1, page 325]\label{4}
Let $T, T_1$ and $T_2$ be topological spaces. Suppose maps $h, h' : T \longrightarrow T_1$ and $k, k' : T_1 \longrightarrow T_2$ are homotopic. Then $k\circ h$ and $k'\circ h'$ are homotopic.
\end{lemma}
\begin{lemma}[\cite{JR}]\label{5}
 Let $T, T_1$ and $T_2$ be topological spaces. Suppose $k : T \longrightarrow T_1$ is a continuous map and  $F$ is a homotopy between maps $f, f' : T_2 \longrightarrow T$, then $k\circ F$ is a homotopy between the maps $k\circ f$ and $k\circ f'$.
\end{lemma}
\begin{definition}
Let $f : T \longrightarrow T'$ be a continuous map. We say that f is a homotopy equivalence if there exists a continuous map $g : T' \longrightarrow T$ such that $f\circ g$ is homotopic to the identity map $I_{T'}$ on $T'$ and  $g\circ f$ is homotopic to the identity map $I_T$ on $T$. Two spaces $T$ and $T'$ are said to be homotopically equivalent or of the same homotopy type if there exists a homotopy equivalence from one to the other.
\end{definition}
\begin{theorem}[\cite{SD}]\label{6}
If two spaces $T$ and $T'$ are homotopically equivalent and any continuous map from $T$ to $S^{n-1}$ is homotopic to a constant map, then  any continuous map from $T'$ to $S^{n-1}$ is homotopic to a constant map. 
\end{theorem}
\begin{theorem}[\cite{SD}]\label{7}
Let $T$ be a simplicial complex of dimension $\leq r$ and $T'$ be a closed subcomplex of $T$. Suppose $f : T' \longrightarrow S^r$ is a continuous map, Then $f$ can be extended to a continuous map $f' : T \longrightarrow S^r$ such that $f'|T' = f$
\end{theorem}
This topological fact can be proved by induction. One chooses a vertex of the complex $T$ that does not belong to $T'$ and chooses an arbitrary extension of $f$. Then one extends $f$ linearly to the edge of $T$ that does not belong to $T'$, then to the two faces etc.
\begin{theorem}[\cite{SD}]\label{8}
Let $T$ be a simplicial complex of dimension $n$ and $T'$ be a closed subcomplex of $T$. Then any continuous map $f : T' \longrightarrow \mathbb{R}^n - \{0\}$ can be extended to a continuous map $f' : T \longrightarrow \mathbb{R}^n$ such that $f'^{-1}(\{0\})$ is a finite set.
\end{theorem}
\begin{theorem}[\cite{HY}, Theorem 4.4, page 153]\label{9}
Let $T$ be a separable metric space and $T'$ be a closed subspace of $T$. Suppose two maps $f$ and $g$ from $T' \longrightarrow S^n$ are homotopic. If there exists an extension $f' : T \longrightarrow S^n$ of $f$, then there also exists an extension $g' : T \longrightarrow S^n$  of $g$ such that $f'$ and $g'$ are homotopic.
\end{theorem}
%\begin{proof}
%Since $f$ and $g$ are homotopic, there exists a continuous map $F : T' \times \mathcal{I} \longrightarrow S^n$ such that $F(x, 0) = f$ and $F(x, 1) = g$. 
%
%Define $H' : (T' \times \mathcal{I})\cup(T \times \{0\}) \longrightarrow S^n$ by 
%\begin{align*}
%H'(x, i) =  \left\{ \begin{array}{rcl} F(x, i) & \mbox{for} & x \in T' \\f'(x) & \mbox{for} & x \in T ~\&~ i = 0 \end{array} \right.
%\end{align*} which is obviously a continuous map. Since $(T' \times \mathcal{I}) \cup (T \times \{0\})$ is a closed subset of $T \times \mathcal{I}, ~ H'$ can be extended to $H : T \times \mathcal{I} \longrightarrow S^n$. Take $g' = H(x, 1) : T \longrightarrow S^n$. Then $g'|T' = g$.  
%\end{proof}
Let $I$ be an ideal of $k[X_1, X_2, \cdots, X_m]$. Then $V(I) = \{(x_1$, $x_2$, $\cdots,$ $x_m) \in k^{m} \mid f(x_1, x_2,\cdots,x_m) = 0$, for every $f\in I\}$. By Hilbert basis theorem, every ideal of $k[X_1, X_2,$ $\cdots$, $X_m]$ is finitely generated, so $V(I)$ is the set of common zeros of finitely many polynomials.

%An algebraic variety is the set of common zeros in $k^{n}$ of a set of polynomials in $k[X_1, X_2, \cdots, X_m]$.

\begin{example}
Let $A = \mathbb{R}[X_1, X_2]$ and $I = (X_{1}^{2} + X_{2}^{2} - 1) \subset A$ be an ideal  , then $V(I) \cap \mathbb{R}^{2} = S^{1}$(real sphere).
\end{example}
%Let $U \subset k^m$ be a nonempty variety. Then $I(U)$ is a prime ideal in \\$k[X_1, X_2, \cdots, X_m]$, so  $\Gamma(U) = k[X_1, X_2, \cdots, X_m] / I(U)$ is a domain and is called the coordinate ring of $U$.
%For any set $U (\ne \phi)$, let $\Im(U, k)$ be the set of all function from $U$ to $k$, $\Im(U, k)$ is a ring with pointwise multiplication and addition. If $U \subset k^m$ is a variety, a function $\phi \in \Im(U, k)$ is called a polynomial function if there is a polynomial $f(X_1, X_2, \cdots, X_m) \in k[X_1, X_2, \cdots, X_m]$ such that $\phi(a_1, a_2, \cdots, a_m) = f(a_1, a_2, \cdots, a_m)$ for all $(a_1, a_2, \cdots, a_m) \in U$. The polynomial functions form a subring of $\Im(U, k)$ containing $k$. Two polynomials $f, ~g$ determine the same function if and only if $(f - g)(a_1, a_2, \cdots, a_m) = 0$ for all $(a_1, a_2, \cdots, a_m) \in U$ i. e $(f - g) \in I(U)$. We may thus identify $\Gamma(U)$ with the subring of $\Im(U, k)$ consisting of all polynomial function on $U$. 
 
 Since $V(I)\cap \mathbb{R}^m$ is the set of common zeros of finitely many polynomials, it is a closed set in the usual Euclidean topology in $\mathbb{R}^{m}$, where $I$ be an ideal of $\mathbb{R}[X_1, X_2, \cdots, X_m]$. 
More generally for any field $k$, there exists a topology on $k^{m}$, where the subsets of the form $V(I)$ are closed. This topology is called the Zariski topology on $k^{m}$. 

In topology Tietze extension theorem (\cite{JR}, Theorem 3.2, page 212) says that 
``Any continuous map of a closed subset of a normal topological space $T$ into the reals $\mathbb{R}$ may be extended to a continuous map of $T$ into $\mathbb{R}$". As an algebraic analogue ``any polynomial function on $V(I)\cap \mathbb{R}^m$ is the restriction of a polynomial function on $\mathbb{R}^m$".  
\begin{remark}\label{10}
 \begin{enumerate}
\item Let $A = \mathbb{R}[X_1, X_2, \cdots, X_m],~ T = Spec(A)$ and  $V(0) = \mathbb{R}^{m}$. Let $\vec{a} = (a_1, a_2, \cdots, a_n)\in A^{n}$. Then we have a continuous map $F_{\vec{a}} : \mathbb{R}^{m} \longrightarrow \mathbb{R}^{n}$ defined as 
$F_{\vec{a}} (x_1, x_2,\cdots, x_m) = (a_1(x_1, x_2, \cdots, x_m), \cdots$, $a_n(x_1, x_2, \cdots, x_m))$, for every $(x_1, x_2, \cdots, x_m) \in \mathbb{R}^{m}$.
  Similarly if $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I,~ T = Spec(A)$ and  $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$, where $I$ is an ideal of a real algebraic variety in $\mathbb{R}[X_1, X_2, \cdots, X_m]$. Then for $\vec{a} = (a_1, a_2, \cdots, a_n) \in A^{n}$, we have a continuous map $F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n}$ defined as $F_{\vec{a}} (x_1, x_2,\cdots, x_m) = (a_1(x_1, x_2,\cdots, x_m)$, $\cdots, a_n(x_1, x_2,\cdots, x_m))$, for every $(x_1, x_2,\cdots, x_m) \in V_{I}(\mathbb{R})$.
The Hilbert Nullstellensatz says that if ``$A = k[X_1, X_2, \cdots, X_m]$, where $k$ is algebraically closed field and $a_1, a_2, \cdots, a_n \in A$, then  $a_1, a_2, \cdots$, $a_n$ have a common zero in $k^{m}$ if and only if the ideal $\langle a_1, a_2, \cdots, a_n \rangle \ne A$''.
Similar to the Hilbert Nullstellensatz if  $A = \mathbb{R}[X_1, X_2, \cdots, X_m] / I$ and  $\vec{a} = (a_1, a_2, \cdots, a_n)\in A^{n}$ is unimodular \textit{i.e.} $\sum_{i = 1}^{n} a_ib_i = 1$ for some $(b_1, b_2, \cdots, b_n)\in A^{n}$, then $a_1, a_2, \cdots, a_n$ do not simultaneously vanish at any point of  $V_{I}(\mathbb{R})$. So we have a map $F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n}-\{0, \cdots, 0\}$.
\item
Now define a map 
$g :\mathbb{R}^{n} - \{(0, 0, \cdots, 0)\} \longrightarrow S^{n-1}$ as $g(x) = \dfrac{x}{||x||}$ for every $x \in \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$, where
$||x||$ denotes norm of $x$. Thus we have a map $G_{\vec{a}} = g\circ F_{\vec{a}}: V_{I}(\mathbb{R}) \longrightarrow S^{n-1} (\subset \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\})$. 
This shows that for any unimodular row $\vec{a} = (a_1, a_2, \cdots, a_n)$ over $A$, we have two continuous maps $F_{\vec{a}}$ and $G_{\vec{a}}$.
\\\noindent \textbf{Claim}: $F_{\vec{a}}$ and $G_{\vec{a}}$ are homotopic. 

Since identity map $Id$ on $\mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ is homotopic to $g$ (straight line homotopy) and $F_{\vec{a}}$ is homotopic to itself, map
$Id \circ F_{\vec{a}} = F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ is homotopic to $G_{\vec{a}} = g\circ F_{\vec{a}}: V_{I}(\mathbb{R})   \longrightarrow S^{n-1}$. This proves the claim.
\item Let $A[X] = (\mathbb{R}[X_1, X_2, \cdots, X_m] / I)[X]$, where $I$ is an ideal of a real algebraic variety in $\mathbb{R}[X_1, X_2, \cdots, X_m]$. Then any element $a \in I$ vanishes on $(x_1, x_2, \cdots, x_m, x)$ for any $x \in \mathbb{R}$ and $(x_1, x_2, \cdots, x_m) \in V_{I}(\mathbb{R})$. This shows that  the real points of the variety corresponding to $A[X]$ is $V_{I}(\mathbb{R})\times\mathbb{R}$, where $V_{I}(\mathbb{R})$ is the set of real points of the variety corresponding to $A$. Therefore any unimodular row $(a_1(X), a_2(X), \cdots, a_n(X)) \in {A[X]}^n$ gives two maps $F_{\vec{a}}[X] : V_{I}(\mathbb{R}) \times\mathbb{R} \longrightarrow \mathbb{R}^{n}-\{0, \cdots, 0\}$ and
$G_{\vec{a}}[X]: V_{I}(\mathbb{R}) \times\mathbb{R} \longrightarrow S^{n-1}$. 
\end{enumerate}
\end{remark}
Throughout this chapter for any unimodular row $\vec{a} = (a_1, a_2, \cdots, a_n),~ F_{\vec{a}}$ and $G_{\vec{a}}$ denote maps from $V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$
and from $V_{I}(\mathbb{R})  \longrightarrow S^{n-1}$, respectively. Similarly for any unimodular row $\vec{a}[X] = (a_1(X), a_2(X), \cdots$, $a_n(X)), ~F_{\vec{a}}[X]$ and $G_{\vec{a}}[X]$ denote maps from $V_{I}(\mathbb{R}) \times \mathbb{R} \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ and from $V_{I}(\mathbb{R}) \times\mathbb{R}   \longrightarrow S^{n-1}$, respectively.
\begin{proposition}\label{11}
 Let $\vec{a} = (a_1, a_2, \cdots, a_n)$ and $\vec{b} = (b_1, b_2, \cdots, b_n)$ be two unimodular rows over $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ such that $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (b_1, b_2, \cdots, b_n)$, where $I$ is an ideal of a real algebraic variety in $\mathbb{R}[X_1, X_2, \cdots, X_m]$.
Then the corresponding mappings $F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ and $F_{\vec{b}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ are homotopic.
\end{proposition}
\begin{proof}
Since $\vec{a} \stackrel {E_n(A)}\sim \vec{b}$, there exists $\sigma = \prod_{i = 1}^{r} E_{ij}(\lambda) \in E_n(A)$ such that $\vec{a} \sigma = \vec{b}$.
Take $\sigma(X) = \prod_{i = 1}^{r} E_{ij}(\lambda X) E_n(A[X])$, so we have $\sigma(0) = I_n$ and $\sigma(1) = \sigma$. Thus $\vec{a}\sigma(0) = \vec{a}I_n = \vec{a}$ and $\vec{a} \sigma(1) = \vec{a} \sigma = \vec{b}$. From Proposition \ref{2}, $(a_1, a_2, \cdots, a_n)\sigma(X)$ is a unimodular row over $A[X]$, from Remark \ref{10} (3), we have map $F_{\vec{a}}[X] :V_{I}(\mathbb{R}) \times \mathbb{R}  \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ such that $F_{\vec{a}}[0] = F_{\vec{a}}$ and $F_{\vec{a}}[1] = F_{\vec{b}}$. Thus the maps $F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ and 
$F_{\vec{b}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{(0, 0, \cdots, 0)\}$ are homotopic via $F_{\vec{a}}[X]$.
\end{proof}
\noindent\textbf{Note}: From Remark \ref{10} (2), maps $F_{\vec{a}}$ and $G_{\vec{a}}$ are homotopic and from Proposition \ref{11}, maps $F_{\vec{a}}$ and $F_{\vec{b}}$ are homotopic. Hence
$G_{\vec{a}}$ and $G_{\vec{b}}$ are also homotopic (\textit{i.e.} $G_{\vec{a}} \stackrel {homo}\sim F_{\vec{a}} \stackrel {homo}\sim F_{\vec{b}} \stackrel {homo}\sim G_{\vec{b}} \Longrightarrow G_{\vec{a}} \stackrel {homo}\sim G_{\vec{b}}$).
% Converse of Proposition \ref{11} says that ``Suppose maps given by two unimodular rows are homotopic. Then these unimodular rows are elementary completable. Therefore one may ask that whether converse is true?

Now we will give underlying topological idea of some results on unimodular rows, which shows that how one can think of the following results from topological point of view.
\begin{lemma}\label{12}
 Let $\vec{a} = (a_1, a_2, \cdots, a_n) \in Um_n(A)$ with $a_1$ being a unit of $A$. Then $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (1, 0, \cdots, 0)$.
\end{lemma}
\noindent \textbf{Underlying topological idea}: Suppose $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$.
 Since $(a_1, a_2, \cdots, a_n) \in A^n$ is a unimodular row over $A$, we have a map $G_{\vec{a}} : V_{I}(\mathbb{R})  \longrightarrow S^{n-1}$.
Since $a_1$ is a unit, $a_1b_1 = 1$ for some $b_1 \in A$ \textit{i.e.} $a_1$ does not vanish at any point of $V_{I}(\mathbb{R})$. Therefore for any element of $\vec{p} = (0, x_2, \cdots, x_n) \in S^{n-1}$,
there does not exist an element $\vec{x} \in V_{I}(\mathbb{R})$ such that $G_{\vec{a}}(\vec{x}) = \vec{p}$. In other words $G_{\vec{a}}$ is not surjective.
So Im$(G_{\vec{a}}) \subset S^{n-1} - \{\vec{p}\}$. Since $S^{n-1} - \{\vec{p}\}$ is contractible, map $G_{\vec{a}}$ is homotopic to a constant map.
\begin{lemma}\label{13}
 Let $\vec{a} = (a_1, a_2, \cdots, a_n) \in A^{n}$ be a unimodular row. Suppose $(a_1, a_2, \cdots$, $a_i)$, for any $i < n$, is unimodular. Then $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (1, 0, \cdots, 0)$.
\end{lemma}
\noindent \textbf{Underlying topological idea}:
Suppose $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$. Since $(a_1, a_2, \cdots, a_n) \in A^n$ is a unimodular row over $A$, we have a map $G_{\vec{a}} : V_{I}(\mathbb{R})  \longrightarrow S^{n-1}$. Since $(a_1, a_2, \cdots, a_i)$ is a unimodular, 
$a_1, a_2, \cdots, a_i$ do not vanish simultaneously at any point of $V_{I}(\mathbb{R})$. Therefore for any element $\vec{q} \in S^{n-1}$, whose first $i$-th coordinates are zero,
there does not exist an element $\vec{x} \in V_{I}(\mathbb{R})$ such that $G_{\vec{a}}(\vec{x}) = \vec{q}$. Hence proof is similar to the proof of Lemma \ref{12}. 

The inclusion map from $S^1$ to $\mathbb{R}^{2} - \{0\}$ is not homotopic to a constant map. As an algebraic consequence we have the following:- 
\begin{example} \label{14}
The unimodular row $\vec{a} = (\textbf{x}_1, \textbf{x}_2) \in A^{2}$, where $A = \mathbb{R}[X_1, X_2]/(X_{1}^{2} + X_{2}^{2} - 1)$ satisfies the property that $(\textbf{x}_1, \textbf{x}_2)$ can not be transformed to $(1, 0)$ via an element of $E_2(A)$ \textit{i.e.} there does
not exist a matrix $\sigma \in E_2(A)$ such that  $(\textbf{x}_1, \textbf{x}_2)\sigma = (1, 0)$. 
\end{example}
\begin{proof} Assume contrary. Since $(\textbf{x}_1, \textbf{x}_2) \in Um_2(A)$ and $V_{(X_{1}^{2} + X_{2}^{2} - 1)}(\mathbb{R}) = V((X_{1}^{2} + X_{2}^{2} - 1)) \cap \mathbb{R}^{2} = S^1$, we have a map 
$F_{\vec{a}} : S^1  \longrightarrow \mathbb{R}^{2} - \{0\}$ which is a inclusion map.
 Suppose there exists $\sigma \in E_2(A)$ such that $(\textbf{x}_1, \textbf{x}_2)\sigma = (1, 0)$. Then it follows $(\textbf{x}_1, \textbf{x}_2)\sigma(X) \in Um_2(A[X])$ (\ref{2}). So we have a map 
$F_{\vec{a}}[X] : S^1 \times\mathbb{R} \longrightarrow \mathbb{R}^{2} - \{0\}$ such that $F_{\vec{a}}[0] = F_{\vec{a}}$ and $F_{\vec{a}}[1] =$ constant map. This shows that inclusion map $F_{\vec{a}}$ is homotopic to a constant map, which is not possible. Hence our assumption is not true.
\end{proof}
\begin{theorem}[\cite{HB}, Theorem 9.3]\label{15}
 Let $A$ be a Noetherian ring of dimension $d$. Let $(a_1, a_2, \cdots$, $a_n) \in A^{n}$ be a unimodular row with $n \geq d+2$. Then $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (1, 0, \cdots, 0)$.
\end{theorem}
\noindent\textbf{Underlying topological idea}:
Let $A$ be the coordinate ring of a real algebraic variety of dimension $d$. Then the dimension of $V_{I}(\mathbb{R}) \leq d$. Since $(a_1, a_2, \cdots, a_n)$ is unimodular, we have a continuous map $G_{\vec{a}}: V_{I}(\mathbb{R})\longrightarrow S^{n-1}$. Assume that $V_{I}(\mathbb{R})$ is a simplicial complex. By simplicial approximation there exists a simplicial map $\psi : V_{I}(\mathbb{R})\longrightarrow S^{n-1}$ such that $\psi$ and $G_{\vec{a}}$ are homotopic via straight line homotopy. 
Since $n \geq d+2$, $n-1 \geq d+1$. Therefore $\psi$ is not surjective (because simplicial map can not raise dimension) \textit{i.e.} Im$(\psi) \subset S^{n-1} - \{\vec{p}\}$. 
Since $S^{n-1} - \{\vec{p}\}$ is contractible, map $\psi$ is homotopic to a constant map. Hence $G_{\vec{a}}$ is also homotopic to a constant map.  
\begin{theorem}\label{16}
  Let $A$ be a Noetherian ring of dimension $d$. Let $(a_1(X), a_2(X)$, $\cdots, a_n(X)) \in A[X]^{n}$ be a unimodular row with $n \geq d+2$. Then
 \begin{align*}
 (a_1(X), a_2(X), \cdots, a_n(X)) \stackrel {E_n(A[X])}\sim (1, 0, \cdots, 0). 
\end{align*} 
\end{theorem}
\noindent\textbf{Underlying topological idea}:
Let $A$ be the co-ordinate ring of a real algebraic variety of dimension $d$. Then the dimension of $V_{I}(\mathbb{R}) \leq d$. Assume that $V_{I}(\mathbb{R})$ is a simplicial complex. For $n \geq d+3$, theorem follows from Theorem \ref{15}. Since $(a_1(X), a_2(X), \cdots, a_n(X)) \in A[X]^{n}$ is unimodular, we have a continuous map $G_{\vec{a}}[X]: V_{I}(\mathbb{R})\times \mathbb{R} \longrightarrow S^{n-1}$. Consider inclusion map $i : V_{I}(\mathbb{R}) \longrightarrow V_{I}(\mathbb{R})\times \mathbb{R}$ and projection $p : V_{I}(\mathbb{R})\times \mathbb{R} \longrightarrow V_{I}(\mathbb{R})$. Then $p\circ i = Id_{V_{I}(\mathbb{R})}$ which is obviously homotopic to identity map on $V_{I}(\mathbb{R})$. On the other hand map $H(x, t) = t(i\circ p)(x) + (1-t)Id_{V_{I}(\mathbb{R})\times \mathbb{R}}$ gives a homotopy between $i\circ p$ and $Id_{V_{I}(\mathbb{R})\times \mathbb{R}}$.

 This shows that spaces $V_{I}(\mathbb{R})$ and $V_{I}(\mathbb{R})\times \mathbb{R}$ are homotopically equivalent. Also from Theorem \ref{15}, any map from $V_{I}(\mathbb{R}) \longrightarrow S^{n-1}$ is homotopic to a constant map. Therefore from Theorem \ref{6}, $G_{\vec{a}}[X]$ is also homotopic to a constant map.
 
 The following theorem is a particular case of the Lemma 4.2.13 (Chapter 4).
\begin{theorem}\label{17}
 Let $A$ be a Noetherian ring of dimension $\leq r$ and $J$ be an ideal of $A$. Let $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_{r+1}}) \in (A/J)^{r+1}$ be a unimodular row.
Then there exist $c_1, c_2, \cdots, c_{r+1}$ such that $\overline{c_i} = \overline{a_i}$ and $(c_1, c_2, \cdots, c_{r+1})$ is unimodular over $A$.
\end{theorem}
 
\noindent\textbf{Underlying topological idea}:
Suppose $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$. Assume that $V_{I}(\mathbb{R})$ is a simplicial complex. Take $V_{J}(\mathbb{R}) = V(J)\cap \mathbb{R}^{m}$. Then $V_{J}(\mathbb{R})$ is a closed subspace of $V_{I}(\mathbb{R})$. Since  $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_{r+1}}) \in (A/J)^{r+1}$ is a unimodular row, we have a continuous map $G_{\vec{a}} : V_{J}(\mathbb{R}) \longrightarrow S^r$. By Theorem \ref{7}, $G_{\vec{a}}$ can be extended to $G'_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow S^r$ such that ${G'_{\vec{a}}}_{|V_{J}(\mathbb{R})} = G_{\vec{a}}$.   

Now we will give motivation about algebraic proof of Theorem \ref{17} by topological proof of Theorem \ref{7}.

Consider Spec($A$) as a simplicial complex with vertices as minimal prime ideals of $A$, edges as height one prime ideals of $A$ and triangles as height $2$ prime ideals etc. We say that an element $a \in A$ vanishes on an edge corresponding to $\mathfrak{p}$ if $a \in \mathfrak{p}$. Also assume $J = \langle a_{r+2} \rangle$.

Now $K = \{\mathfrak{p} \in Spec(A)~|~ a_{r+2} \in \mathfrak{p}\}$ is a sub-complex of Spec($A$) and we have a row $(a_1, a_2, \cdots, a_{r+1})$ which does not vanish on $K$.
 
 Now we choose vertices of Spec($A$) which do not belong to $K$ \textit{i.e.} choose minimal prime ideals $\mathfrak{p}_1, \mathfrak{p}_2, \cdots, \mathfrak{p}_s$ of $A$ such that $a_{r+2} \notin \mathfrak{p}_i, ~ 1 \leq i \leq s$. Thus $\langle a_1, a_{r+2} \rangle \notin \mathfrak{p}_i, ~ 1 \leq i \leq s$. Therefore there exists $\lambda_1 \in A$ such that  $a'_1 = a_1 + \lambda_1 a_{r+2} \notin \cup_{i=1}^{s}\mathfrak{p}_i$. Hence we have a row $(a'_1 , a_2, \cdots, a_{r+1})$ which does not vanish on $K'$, where $K' = K \cup (\cup_{i=1}^{s}\mathfrak{p}_i)$.
 
Now we choose edges  of Spec($A$) which do not belong to $K'$ \textit{i.e.} choose prime ideals $\mathfrak{q}_1, \mathfrak{q}_2, \cdots, \mathfrak{q}_l$ containing $a'_1$ but $a_{r+2} \notin \mathfrak{q}_i$. If no such prime ideal with the above property exists, then $(a'_1, a_2, \cdots, a_{r+1})$ does not vanish on Spec($A$). So we are done. Otherwise, since $a_{r+2} \notin \mathfrak{q}_i$, $\langle a_2, a_{r+2} \rangle \nsubseteq \mathfrak{q}_i,~ 1 \leq i \leq l \Rightarrow \langle a_2, a_{r+2} \rangle \nsubseteq \cup_{i=1}^{l}\mathfrak{q}_i$. So there exists $\lambda_2$ such that $a'_2 = a_2 + \lambda_2 a_{r+2} \notin \cup_{i=1}^{l}\mathfrak{q}_i$. Hence we have a row $(a'_1 , a'_2, \cdots, a_{r+1})$ which does not vanish on $K''$, where $K'' = K' \cup (\cup_{i=1}^{l}\mathfrak{q}_i)$. Continuing same procedure we get the results.

The following theorem is a particular case of the Lemma 4.2.5 (Chapter 4).
\begin{theorem}\label{18}
 Let $A$ be a Noetherian ring of dimension $n$ and $J$ be an ideal of $A$. Suppose $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_n}) \in (A/J)^n$ is a unimodular row.
 Then there exist $b_1, b_2, \cdots, b_n \in A$ such that $\overline{b_i} = \overline{a_i}$, for all $i$ and ideal $\langle b_1, b_ 2,\cdots, b_n \rangle$ has height $n$.
% (that is $b_1, b_ 2,\cdots, b_n$ have finitely many zeros).
\end{theorem}
\noindent\textbf{Underlying topological idea}:
Suppose $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$. Assume that $V_{I}(\mathbb{R})$ is a simplicial complex. Take $V_{J}(\mathbb{R}) = V(J)\cap \mathbb{R}^{m}$. Then $V_{J}(\mathbb{R})$ is a closed subspace of $V_{J}(\mathbb{R})$. Since  $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_{r+1}}) \in (A/J)^{r+1}$ is a unimodular row, we have a continuous map $G_{\vec{a}} : V_{J}(\mathbb{R}) \longrightarrow S^r$. By Theorem \ref{8}, $G_{\vec{a}}$ can be extended to a continuous  map $G'_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow S^r$ such that ${G'_{\vec{a}}}_{|V_{J}(\mathbb{R})} = G_{\vec{a}}$. 
\begin{lemma}\label{19}
 The canonical homomorphism of groups from $E_n(A)$ to $E_n(A/I)$ is surjective.
\end{lemma}
The lemma follows from the fact that generators $E_{ij}(\overline{\lambda})$ of $E_n(A/I)$ can be lifted to generators $E_{ij}(\lambda)$ of $E_n(A)$. 

We show by an example that the canonical homomorphism from $SL_n(A)$ to $SL_n(A/I)$ need not be surjective.
\begin{example}\label{20}
 Let $B = \mathbb{R}[X, Y], A = \mathbb{R}[X, Y] / (X^{2} + Y^{2} -1)$. Then the unimodular row $(\textbf{x}, \textbf{y}) \in A^{2}$ can not be lifted to a unimodular row over $B^{2}$.
\end{example}
\begin{proof}
Let $ \alpha = \left(\begin{array}{cccc} \textbf{x} & \textbf{y} \\ -\textbf{y} & \textbf{x}\end{array}\right) \in SL_2(A)$.

\noindent Claim: There does not exist $\beta \in SL_2(B)$ such that $\overline{\beta} = \alpha$.\\ Since $\alpha \in A$, we have a map $\phi : S^1 \longrightarrow  SL_2(\mathbb{R})$ defined by 
\begin{align*}
\phi(x_1, x_2) = \alpha(x_1, x_2) = \left(\begin{array}{cccc} \textbf{x}(x_1, x_2) & \textbf{y}(x_1, x_2)\\ -\textbf{y}(x_1, x_2) & \textbf{x}(x_1, x_2) \end{array}\right) = \left(\begin{array}{cccc} x_1 & x_2\\ -x_2 & x_1 \end{array}\right).
\end{align*} Let 
\begin{align*}
\beta = \left(\begin{array}{cccc} f_1(X, Y) & f_2(X, Y)\\ -f_3(X, Y) & f_4(X, Y) \end{array}\right)\in SL_2(B)
\end{align*} be a lift of $\alpha$.
Therefore we have a map $\Phi : \mathbb{R}^2 \longrightarrow SL_2(\mathbb{R})$ defined by
\begin{align*}
\Phi(x_1, x_2) = \beta(x_1, x_2) =  \left(\begin{array}{cccc} f_1(x_1, x_2) & f_2(x_1, x_2)\\ -f_3(x_1, x_2) & f_4(x_1, x_2) \end{array}\right)
\end{align*} which is clearly an extension of $\phi$. In particular considering the first row of $\alpha ~\&~ \beta$ we see that inclusion map from $S^1 \longrightarrow \mathbb{R}^2 - \{0\}$ extends to $\mathbb{R}^2 \longrightarrow \mathbb{R}^2 - \{0\}$ which is not possible. Hence claim is proved. Thus the unimodular row $(\textbf{x}, \textbf{y}) \in A^{2}$ can not be lifted to a unimodular row over $B^{2}$.\\
\textbf{Note}:  From Example \ref{20}, it is clear that $\left(\begin{array}{cccc} \textbf{x} & \textbf{y} \\ -\textbf{y} & \textbf{x}\end{array}\right)$ does not belong to $E_2(A)$ otherwise it could be lifted to a matrix in $E_2(B) \subset SL_2(B)$ \textit{i.e.} the unimodular row $(\textbf{x}, \textbf{y}) \in A^{2}$ is not elementary completable.
 \end{proof}
The general from of the Lemma \ref{19} is the following fact-
\begin{theorem}\label{21}
 Let $\vec{a} = (a_1, a_2, \cdots, a_n) \in A^{n}$ be a unimodular row. Suppose $J$ is an ideal of $A$ and $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_n}) \stackrel {E_n(\overline{A})}\sim 
(\overline{b_1}, \overline{b_2}, \cdots, \overline{b_n})$, where bar denotes reduction modulo $J$. Then there exists $(c_1, c_2, \cdots, c_n) \in A^{n}$ such that $(a_1, a_2, \cdots, a_n) \stackrel {E_n(A)}\sim (c_1, c_2, \cdots, c_n)$ and $(\overline{c_1}, \overline{c_2}, \cdots, \overline{c_n}) = (\overline{b_1}, \overline{b_2}, \cdots, \overline{b_n})$. In particular  $(\overline{b_1}, \overline{b_2}, \cdots, \overline{b_n})$ can be lifted to a unimodular row over $A$.
\end{theorem}
\noindent\textbf{Underlying topological idea}:
 Suppose $A = \mathbb{R}[X_1, X_2, \cdots, X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$. Assume that $V_{I}(\mathbb{R})$ is a separable metric space. Take $V_{J}(\mathbb{R}) = V(J)\cap \mathbb{R}^{m}$. Then $V_{J}(\mathbb{R})$ is a closed subspace of $V_{I}(\mathbb{R})$.
Since $(\overline{a_1}, \overline{a_2}, \cdots, \overline{a_n}) \stackrel {\overline{\alpha} \in E_n(\overline{A})}\sim (\overline{b_1}, \overline{b_2}, \cdots, \overline{b_n})$, by Proposition \ref{11}, the corresponding continuous maps $G'_{\vec{a}} : V_{J}(\mathbb{R}) \longrightarrow S^{n-1}$ and $G'_{\vec{b}} : V_{J}(\mathbb{R}) \longrightarrow S^{n-1}$ are homotopic. That is there exists a continuous map $F : V_{J}(\mathbb{R}) \times \mathcal{I} \longrightarrow S^{n-1}$ such that $F(x, 0) = G'_{\vec{a}}$ and $F(x, 1) = G'_{\vec{b}}$.

Let $\alpha$ be the lift of $\overline{\alpha}$. Take $(c_1, c_2, \cdots, c_n) = (a_1, a_2, \cdots, a_n)\alpha$. Thus we have an extension $G_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow S^{n-1}$ given by $(c_1, c_2, \cdots, c_n)$ of $G'_{\vec{a}}$. \\Define $H' : (V_{J}(\mathbb{R}) \times \mathcal{I})\cup(V_{I}(\mathbb{R}) \times \{0\}) \longrightarrow S^n$ by 
\begin{align*}
H'(x, t) = \left\{\begin{array}{rcl}F(x, t) & \mbox{for} & x \in V_{J}(\mathbb{R}) ~\&~ 0 \leq t \leq 1  \\ G_{\vec{a}}& \mbox{for} & x \in V_{I}(\mathbb{R}) ~\&~ t = 0 \end{array}\right.
\end{align*} which is obviously a continuous map. Since $(V_{J}(\mathbb{R}) \times \mathcal{I}) \cup (V_{I}(\mathbb{R}) \times \{0\})$ is a closed subset of $V_{I}(\mathbb{R}) \times \mathcal{I}$, $H'$ can be extended to $H : V_{I}(\mathbb{R}) \times \mathcal{I} \longrightarrow S^n$ (from proof of Theorem \ref{9}). Take $G_{\vec{b}} = H(x, 1) : V_{I}(\mathbb{R}) \longrightarrow S^n$. Then ${G_{\vec{b}}}_{|V_{J}(\mathbb{R})} = G'_{\vec{b}}$. 
\begin{example}
Let $A = \mathbb{R}[X, Y, Z]/(X^{2}+Y^{2}+Z^{2}-1)$. Then $(\textbf{x}, \textbf{y}, \textbf{z})\in A^{3}$ is a unimodular row. Since the identity map from $S^{2}$ to itself is not homotopic 
to a constant map, we have as an algebraic consequence that $(x, y, z)$ is not equivalent to $(1, 0, 0)$ via the action of $E_3(A)$. In other words $(\textbf{x}, \textbf{y}, \textbf{z})$ is not completable
to an elementary matrix. In fact $(\textbf{x}, \textbf{y}, \textbf{z})$ is not completable to a matrix in $GL_3(A)$.
\end{example}
\begin{proof}
 Assume contrary that $(\textbf{x}, \textbf{y}, \textbf{z})$ is the first row of a matrix in $GL_3(A)$ \textit{i.e.} $(\textbf{x}, \textbf{y}, \textbf{z})$ is a completable unimodular row. In other words $P \cong A^{3}/\langle \textbf{x}, \textbf{y}, \textbf{z} \rangle \cong  A^{2}$. Thus we have a surjective homomorphism $f : P \longrightarrow A$. Suppose $e_1, e_2, e_3$ are the standard basis vectors of $A^{3}$ and $f(\overline{e_i}) = h_i$ for $i = 1, 2, 3$. Since $f(\textbf{x}, \textbf{y}, \textbf{z}) = 0$,  we have $\textbf{x}f(\overline{e_1}) + \textbf{y}f(\overline{e_2}) + \textbf{z}f(\overline{e_3}) = 0 \Rightarrow \textbf{x}h_1 + \textbf{y}h_2 + \textbf{z}h_3 = 0 $. This implies that $Xh_1(X, Y, Z) + Yh_2(X, Y, Z) + Zh_3(X, Y, Z)$ is a multiple of $X^{2}+Y^{2}+Z^{2}-1$.

In particular, if $x_{1}^{2}+x_{2}^{2}+x_{3}^{2} = 1$, then $x_{1}h_1(x_1, x_2, x_3) + x_{2}h_2(x_1, x_2, x_3) + x_{3}h_3(x_1, x_2, x_3)$ = $0$ implies that $(h_1(x_1, x_2, x_3)$, $h_2(x_1, x_2, x_3)$, $h_3(x_1, x_2, x_3))$ is perpendicular to $(x_1, x_2, x_3)$. Define a continuous vector field $\Phi_1 : S^{2} \longrightarrow \mathbb{R}^{3}$ by $\Phi_1(x_1, x_2, x_3)= (h_1(x_1, x_2, x_3)$, $h_2(x_1, x_2, x_3)$, $h_3(x_1, x_2, x_3))$. The zeros of this vector field are those point $(x_1, x_2, x_3) \in S^2$ where $h_1(x_1, x_2, x_3)$, $h_2(x_1, x_2, x_3)$, $h_3(x_1, x_2, x_3)$ are all zero.
 Since $f(\overline{e_i}) = h_i$, the ideal $(h_1(\textbf{x}, \textbf{y}, \textbf{z}), h_2(\textbf{x}, \textbf{y}, \textbf{z})$, $h_3(\textbf{x}, \textbf{y}, \textbf{z})) = A$. Hence the corresponding vector field on $S^2$ has no real zeros, contradicting the fact that there is no nowhere vanishing continuous vector field on $S^2$.

Thus $(\textbf{x}, \textbf{y}, \textbf{z})$ is not completable to a
 matrix in $GL_3(A)$ implies that $(\textbf{x}, \textbf{y}, \textbf{z})$ is not completable to a matrix in $E_3(A)$.  
\end{proof}
\begin{proposition}[\cite{MKR}, Corollary 2.2]\label{221}
Let $(a, b, c) \in A^3$ be a unimodular row. Then $(a^2, b, c)$ is completable.
\end{proposition}
To understood underlying topological idea, we first give a proof of Proposition \ref{221}.
\begin{proof}
Since $(a, b, c)$ is a unimodular, there exist $a', b' ~\&~ c'$ such that $aa' + bb' + cc' = 1$. Consider the matrix
\begin{align*}
\alpha = \left(\begin{array}{cccc} 0 & a & b & c\\ -a & 0 & c' & -b'\\ -b & -c' & 0 & a' \\-c & b' & -a' & 0\end{array}\right).
\end{align*} Then $det(\alpha) = (aa' + bb' + cc')^2 = 1$.
Since $\langle 0, -a, -b, -c \rangle = A, ~ (0, -a, -b, -c)$ is a elementary completable. The matrix 
$\alpha_{1}\alpha_2$ is a completion of $(0, -a, -b, -c)$, where 
\begin{align*}
\alpha_1 = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ a & 1 & 0 & 0\\ b & 0 & 1 & 0 \\ c & 0 & 0 & 1\end{array}\right) 
\end{align*}
and 
\begin{align*}
\alpha_2 = \left(\begin{array}{cccc} 1 & -a' & -b' & -c'\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right). 
\end{align*}
\vspace{-1cm}
Take $\alpha' = \alpha_{1}\alpha_{2}\alpha$. Then 
\begin{align*}
\alpha' = \left(\begin{array}{cccc} 1 & a & b & c\\ 0 & a^2 & ab + c' & ac - b'\\ 0 & ab-c' & b^2 & bc + a' \\ 0 & ac + b' & bc - a' & c^2\end{array}\right)
\end{align*}
 and $det(\alpha') = det(\alpha)$. By replacing $a \rightarrow a, b \rightarrow b, c \rightarrow c, a' \rightarrow a', b' \rightarrow b'+ac, c' \rightarrow  c'-ab$, we have
 \begin{align*}
 \beta = \left(\begin{array}{cccc} 1 & a & b & c\\ 0 & a^2 & c' & -b'\\ 0 & 2ab-c' & b^2 & bc+a' \\ 0 & 2ac+b' & bc-a' & c^2\end{array}\right).
 \end{align*}
 Then $det(\beta) = aa'+b(b'+ac)+c(c'-ab) = aa' + bb' + cc' = 1$. Again by replacing $a \rightarrow a, b \rightarrow -b', c \rightarrow c', a' \rightarrow a', b' \rightarrow -b, c' \rightarrow c$,
we have 
\begin{align*}
\sigma = \left(\begin{array}{cccc} 1 & a & -b' & c'\\ 0 & a^2 & c & b\\ 0 & -2ab'-c & b'^2 & -b'c'+a' \\ 0 & 2ac'-b & -b'c'-a' & c'^2\end{array}\right).
\end{align*}
Then $det(\sigma) = aa' + (-b')(-b+ac') + c'(c+ab') = aa' + bb' + cc' = 1$. It is clear that $det(\sigma) = det \left(\begin{array}{cccc}  a^2 & c & b\\ -2ab'-c & b'^2 & -b'c'+a' \\ 2ac'-b & -b'c'-a' & c'^2\end{array}\right) = 1$. Hence the matrix
\begin{align*}
\alpha_{3} = \left(\begin{array}{cccc}  a^2 & b & c\\ -2ac'-b & c'^2 & -b'c'+a' \\ 2ab'-c & -b'c'-a' & b'^2\end{array}\right) 
\end{align*} is a completion of $(a^2, b, c)$.
\end{proof}
\noindent\textbf{Underlying topological idea:}
Let $A = \mathbb{R}[X_1, X_2, \cdots , X_m]/I$ and $V_{I}(\mathbb{R}) = V(I) \cap \mathbb{R}^{m}$.
 Since $(a, b, c) \in A^3$ is a unimodular row, $(0, -a, -b, -c)$ is elementary completable. \\ 
Consider an exact sequence $0 \longrightarrow  SL_3(\mathbb{R}) \overset{i_1}\longrightarrow SL_4(\mathbb{R}) \overset{i_2} \longrightarrow \mathbb{R}^{4} - \{0\} \longrightarrow 0$, where $i_1$ sends any matrix $\sigma' \in SL_3(\mathbb{R})$ to $\left(\begin{array}{cccc}  1 & 0\\ 0 & \sigma' \end{array}\right) \in SL_4(\mathbb{R})$ and  $i_2$ sends any matrix $\sigma_1 \in SL_4(\mathbb{R})$ to its first row.

Since $\alpha \in SL_4(A)$, we have a map $\phi_1 : V_{I}(\mathbb{R}) \longrightarrow SL_4(\mathbb{R})$
defined as $\phi_1(\vec{x}) = \alpha(\vec{x})$ and for any unimodular row of length $4$, we have a map from $V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{4} - \{0\}$. The map $i_2 \circ \phi_1 : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{4} - \{0\}$ is equal to the map given by unimodular row $(0, -a, -b, -c)$ which is homotopic to constant map (by Proposition \ref{11}). Also $\alpha_3 \in SL_3(A)$ gives a map $\phi_2 : V_{I}(\mathbb{R}) \longrightarrow SL_3(\mathbb{R})$ such that $i_1 \circ \phi_2$ homotopic to the map $\phi_1$ and the map $i'_2 \circ \phi_2 : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{3} - \{0\}$ is equal to the map given by unimodular row $(a^2, b, c)$, where $i'_2$ sends any matrix $\sigma_1 \in SL_3(\mathbb{R})$ to its first row. 
\section{On a lemma of Vaserstein's}
Throughout this section $T$ is a compact Hausdorff topological space (\textit{i.e.} normal space), $C(T)$ is the ring of real valued continuous functions on $T$ and $v_0 \stackrel {GL_n(C(T))}\sim v_t$ means there exists a matrix $\alpha \in GL_n(C(T))$ such that $v_0 \alpha = v_t$ , where $v_0$ and $v_t$ are unimodular row in $C(T)$. This is an equivalence relation.

We now give a proof of a result which says that if there is no nowhere vanishing continuous vector field on $S^2$, then $S^2$ is not contractible. This proof is motivated from Simha (\cite{RS}).

To begin proof we need some preliminaries on reflections:-

Let $w (\ne 0) \in \mathbb{R}^n$ be a vector. A reflection about $w$ is a linear transformation $\sigma : \mathbb{R}^n \longrightarrow \mathbb{R}^n$ which satisfies $\sigma(w) = -w, ~\&~ \sigma(w_1) = w_1$,
 where $w_1 \in w^{\bot} = \{ w'\in \mathbb{R}^n ~|~ \langle w', w \rangle = 0\}$. 

Let $v \in \mathbb{R}^n$ such that $v = v_1 + \lambda w, v_1 \in w^{\bot}$. Then $\sigma (v) = v_1 - \lambda w$. We have $\langle v, w \rangle = \langle v_1, w \rangle + \lambda \langle w, w \rangle = \lambda \langle w, w \rangle$, since $v_1 \in w^{\bot}$. This implies that 
$\lambda = \dfrac{\langle v, w \rangle}{\langle w, w \rangle}$. Therefore $\sigma(v) = v_1 -  \lambda w = v_1 + \lambda w - 2 \lambda w  = v - 2 \lambda w  = v - 2 \dfrac{\langle v, w \rangle}{\langle w, w \rangle} w$. This map $\sigma$ is denoted by $\sigma _w$. 

If $v_1$ and $w_1$ are two vectors in $\mathbb{R}^{n}$ and $||v_1|| = ||w_1||$, then we have a rhombus whose sides are $v_1, w_1$ and whose diagonal
are $v_1 + w_1, v_1 - w_1$. Thus $\sigma_{v_1 - w_1}(v_1)$ = $v_1 -2 \dfrac{\langle v_1, v_1 - w_1 \rangle}{\langle v_1 - w_1, v_1 - w_1 \rangle} (v_1 - w_1)$ = $v_1 - 
2 \dfrac{||v_1||^{2} - \langle v_1, w_1 \rangle}{||v_1||^{2} - 2\langle v_1, w_1 \rangle + ||w_1||^{2}} (v_1 - w_1)$  = $v_1 - (v_1 - w_1) = w_1$. Also we have $\sigma_{v_1 + w_1}(v_1) = -w_1$. 
\begin{remark}\label{22}
Any continuous map from $T \longrightarrow \mathbb{R}^n - \{0\}$ leads to a unimodular row $(a_1, a_2, \cdots$, $a_n)$ over the ring of continuous function $C(T)$, where $a_i$ is the projection from $T \longrightarrow \mathbb{R}$  because the element $\sum a_{i}^{2} \in  \langle a_1, a_2, \cdots, a_n \rangle$ is a continuous function on $T$ which does not vanish at any point of $T$ otherwise each $a_i$ will vanish at that point. In particular, any continuous map $T \longrightarrow S^{n-1}$ gives rise to a unimodular row over $C(T)$.  Let $H : T \times \mathcal{I} \longrightarrow S^{n-1}$ be a continuous map.\\
\noindent\textbf{Claim}: For sufficiently small $t$, $v_0 \stackrel {GL_n(C(T))}\sim v_t$, where $v_0$ and $v_t$ are the corresponding unimodular rows given by the maps $H(x, 0) : T \longrightarrow S^{n-1}$ and  $H(x, t) : T \longrightarrow S^{n-1}$, respectively. 
\begin{proof}
By continuity of $H$ and compactness of $T$, it follows that there exists a $t_\theta > 0$ such that for $t < t_\theta$, $v_0$ and $v_t$ are sufficiently close in the sense that $v_0(p)$ and $v_t(p)$ are not antipodal for every $p$ \textit{i.e.} $v_0(p) + v_t(p) \ne 0$ for $t < t_\theta$ and for every $p\in T$. Then for every $p \in T, \sigma_{v_0(p) + v_t(p)}$ is a reflection which is an element of $O_n(\mathbb{R})$ and satisfies $\sigma_{v_0(p) + v_t(P)}(v_0(p)) = -v_t(p)$. Hence there exists $\alpha : T \longrightarrow O_n(\mathbb{R})$ sending $p$ to $-\sigma_{v_0(p) + v_t(p)}$ such that $\alpha v_0 = v_t$. Since $O_n(\mathbb{R}) \subset GL_n(\mathbb{R}), \sigma \in GL_n(C(T))$ and $v_0 \stackrel {GL_n(C(T))}\sim v_t$.
\end{proof}
 \end{remark}
 Now we will give a proof of the fact that $S^2$ is not contractible.
\begin{proof} Assume contrary that the real two sphere $S^2$ is contractible. Then there exists a continuous map $H : S^{2} \times \mathcal{I} \longrightarrow S^2$, where $\mathcal{I} = [0, 1]$
such that the map $H(x, 0)$ is a constant map given by $(1, 0, 0)$ and $H(x, 1)$ is a identity map on $S^2$ corresponding to the unimodular row $(\textbf{x}, \textbf{y}, \textbf{z}) \in (C(S^2))^3$.


Now let $S = \{ t \in \mathcal{I} \mid v_0 \stackrel {GL_n(C(T))}\sim v_t\}$. By the compactness and connectedness of $\mathcal{I}$,  it is easy to see that $S = \mathcal{I}$ \textit{i.e.} $v_0 \stackrel {GL_n(C(T))}\sim v_1$. Hence from the claim it is clear that $(1, 0, 0)\stackrel {GL_3(C(S^2))}\sim (\textbf{x}, \textbf{y}, \textbf{z})$ \textit{i.e.} $(\textbf{x}, \textbf{y}, \textbf{z})$ is completable which is not possible. Hence $S^2$ is not contractible.
\end{proof}
In Section 5.2, we have seen  that if $A$ is the co-ordinate ring of a real algebraic variety and $\vec{a} \in A^n$ is unimodular, then $\vec{a}$ gives rise to a continuous function $F_{\vec{a}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{0\}$. Also if $\vec{b} \in A^{n}$ is unimodular and $\vec{a} \stackrel {E_n(A)}\sim \vec{b}$, then the corresponding maps $F_{\vec{a}}, F_{\vec{b}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n} - \{0\}$ are homotopic.

We shall now investigate the extent to which the converse is valid. Simha's proof (\cite{RS}) shows that 
if $T$ is a compact topological space and $G_{\vec{a}}, G_{\vec{b}} : V_{I}(\mathbb{R}) \longrightarrow S^{n-1}$ are continuous maps which are homotopic, then the corresponding unimodular rows $\vec{a} ~\&~ \vec{b}$ satisfy $\vec{a} \stackrel {GL_n(C(T))}\sim \vec{b}$.
To investigate the converse, we need a lemma of Vaserstein:-
\begin{lemma}[\cite{VS}]\label{23}
Let $A$ be a ring and $\vec{a} = (a_1, a_2, \cdots, a_n) \in A^{n}$ be a unimodular row. Let $\vec{b} = (b_1, b_2, \cdots, b_n)$ and $\vec{c} = (c_1, c_2, \cdots, c_n)$ be such that 
$\sum_{i = 1}^{n} a_ib_i = \sum_{i = 1}^{n} a_ic_i = 1$. Then there exists a matrix $\alpha \in SL_n(A)$ which can be connected to the identity matrix such that $\vec{b} \stackrel {\alpha}\sim \vec{c}$, if $n \geq 3$.
\end{lemma}
\noindent\textbf{Underlying topological idea}: If $A$ is the co-ordinate ring of a real algebraic variety and $\vec{a}, \vec{b}, \vec{c} \in A^n$ are unimodular rows satisfying the property $\vec{a}\vec{b}^t = \vec{a}\vec{c}^t = 1$. Then we have two continuous maps $F_{\vec{b}}, F_{\vec{c}} : V_{I}(\mathbb{R}) \longrightarrow \mathbb{R}^{n}-\{0\}$.\\
\noindent\textbf{Claim}: $F_{\vec{b}} ~\&~ F_{\vec{c}}$ are homotopic.\\
Since $\vec{a}\vec{b}^t = \vec{a}\vec{c}^t = 1$, for each point $p \in V_{I}(\mathbb{R})$, the vectors $F_{\vec{b}}(p)$ and $F_{\vec{c}}(p)$ are not in antipodal directions (if $F_{\vec{b}}(p) = - F_{\vec{c}}(p)$ means $\langle \vec{a}, \vec{b} \rangle = 1 = - \langle \vec{a}, \vec{c} \rangle$ which is not true). Therefore $H = (1-t)F_{\vec{b}} + tF_{\vec{c}}$ is a continuous map from $V_{I}(\mathbb{R})\times \mathcal{I}$ to $\mathbb{R}^{n}-\{0\}$, which is a homotopy between $F_{\vec{b}} ~\&~ F_{\vec{c}}$. Thus the claim is proved.

To give algebraic proof of Vaserstein's lemma one needs to compute the determinant of matrices of the kind $I_n + \alpha$, where $\alpha$ is a $n \times n$ matrix of rank $1$. Note that a general $n \times n$ matrix of rank $\leq 1$ over a field looks like $x^{t}y$, where $x = (x_1, x_2, \cdots, x_n)$ and $y = (y_1, y_2, \cdots, y_n)$. In particular a $2 \times 2$ matrix of rank $\leq 1$ over a field looks like $x^{t}y = \left(\begin{array}{cccc} x_1y_1 & x_1y_2 \\ x_2y_1 & x_2y_2\end{array}\right)$, where $x = (x_1, x_2)$ and $y = (y_1, y_2)$.
 
For simplicity we state and prove the lemma in the $2 \times 2$ case (the general case being similar).

% Note that a general matrix of rank $\leq 1$ over a field looks like $x^{t}y = \left(\begin{array}{cccc} x_1y_1 & x_1y_2 \\ x_2y_1 & x_2y_2\end{array}\right)$, where $x = (x_1, x_2)$ and $y = (y_1, y_2)$.
 \begin{lemma}\label{24}
 \begin{align*}
det\left(\begin{array}{cccc} 1 + x_1y_1 & x_1y_2 \\ x_2y_1 & 1 + x_2y_2\end{array}\right) = 1 + x_1y_1 + x_2y_2. 
 \end{align*}  
\end{lemma} \textit{i.e.} $det(I_n + x^{t}y) = 1 + xy^{t}$
 \begin{proof}
 Note that
 \begin{align*}
 det\left(\begin{array}{cccc} 1 + x_1y_1 & x_1y_2 & 0\\ x_2y_1 & 1 + x_2y_2 & 0 \\ 0 & 0 & 1 \end{array}\right) = 
 det\left(\begin{array}{cccc} 1 + x_1y_1 & x_1y_2 & x_1\\ x_2y_1 & 1 + x_2y_2 & x_2 \\ 0 & 0 & 1 \end{array}\right) \\ = 
 det\left(\begin{array}{cccc} 1  & 0 & x_1\\ 0 & 1  & x_2 \\ -y_1 & -y_2 & 1 \end{array}\right) = 
 det\left(\begin{array}{cccc} 1 & 0 & x_1\\ 0 & 1 & x_2 \\ 0 & 0 & 1 + x_1y_1 + x_2y_2 \end{array}\right) = 1 + x_1y_1 + x_2y_2. 
\end{align*}
 Since 
 \begin{align*}
det\left(\begin{array}{cccc} 1 + x_1y_1 & x_1y_2 \\ x_2y_1 & 1 + x_2y_2\end{array}\right) = 
 det\left(\begin{array}{cccc} 1 + x_1y_1 & x_1y_2 & 0\\ x_2y_1 & 1 + x_2y_2 & 0 \\ 0 & 0 & 1 \end{array}\right).
 \end{align*} This completes proof of the lemma.
 \end{proof}
\begin{remark}\label{25}
 \begin{enumerate}
 \item The elementary matrices $E_{ij}(\lambda)$ are of the form $I_n + \alpha$, where $\alpha$ has rank $1$. 
 \item The matrix $E_{ij}(1) = I_n + e_{i}^t e_j, i \ne j$. If we conjugate $E_{ij}(1)$ by a matrix $\alpha \in GL_n(k), k$ a field, we obtain the matrix $\alpha E_{ij}(1)\alpha^{-1} = I_n + v^{t}w$, where $v^{t} = \alpha e_{i}^t, w = e_j\alpha^{-1}$. Since $e_ie_j^{t} = 0, vw^{t} = 0$ and by Lemma \ref{24}, it follows that the determinant of the matrix $I_n + v^{t}w$ is $1$.
\item Let $\sigma : k^{n} \longrightarrow k$ be a linear transformation and $w \in k^n$ is a non-zero vector such that $\sigma(w) \ne 0$. Define $\sigma' : k^{n} \longrightarrow k^{n}$ by $\sigma'(v) = v + \sigma(v)w$. Then $\sigma'$ is a linear transformation and its matrix is of the form $I_n + \alpha$, where $\alpha$ has rank $1$.
 Since dim $Ker(\sigma) = n-1$, we can choose a basis $x_1, x_2, \cdots, x_{n-1}$ of $Ker(\sigma)$ and therefore $x_1, x_2, \cdots, x_{n-1}, w$ is a basis of $k^n$. Thus the determinant of $\sigma'$ with respect to the above basis is seen to be $1 + \sigma(w)$.
In particular we can compute the determinant of the reflection transformation which sends $v$ to $v - 2 \dfrac{\langle v, w \rangle}{\langle w, w \rangle} w$. Note 
that in this case $\sigma(v) = - 2 \dfrac{\langle v, w \rangle}{\langle w, w \rangle}$, in particular $\sigma(w) = -2$. The determinant of the reflection transformation is $-1$.
\item If $\sigma(w) = 0$, then we have a basis $v_1, v_2, \cdots, v_{n-1}$ with $v_1 = w$ of $Ker(\sigma)$. Let $v_n \in k^n$ but $v_n \notin Ker(\sigma)$. The determinant
of the matrix corresponding to $\sigma'$ with respect to the basis $v_1, \cdots, v_{n-1}, v_n$ is $1$.
 In particular, it follows as we have seen before that the $det(I_n + v^{t}w) = 1$, where $wv^{t} = 0$.
 \end{enumerate}
\end{remark}
Now we prove Lemma \ref{23} of Vaserstein.
\begin{proof}
Suppose $\vec{a} \in A^{n}$ is unimodular and $\vec{b}, \vec{c} \in A^{n}$ are such that $\vec{a}\vec{b}^t = \vec{a}\vec{c}^t = 1$. Then $\vec{a}(\vec{c} - \vec{b})^{t} = 0$. So we have a matrix $\alpha = I_n + (\vec{c} - \vec{b})^{t}\vec{a}$ satisfying the property that $det(\alpha) = 1$ and $\alpha \vec{b}^{t} = \vec{b}^{t} + (\vec{c} - \vec{b})^t = \vec{c}^t$.
 Take $\beta =  I_n + (\vec{c} - \vec{b})^{t}\vec{a}X \in SL_n(A[X])$. Then $\beta(0) = I_n, \beta(1) = \alpha$ proving the lemma.
\end{proof}
\begin{theorem}
Let $H : T \times \mathcal{I} \longrightarrow S^{n-1}$ be a continuous homotopy such that $v_0$ and $v_1$ are the corresponding unimodular rows given by maps $H(x, 0) : T \longrightarrow S^{n-1}$ and  $H(x, 1) : T \longrightarrow S^{n-1}$, respectively. Then there exists a matrix $\alpha \in SL_n(C(T))$ such that $\alpha$ can be connected to the identity and $v_0 \stackrel {\alpha}\sim v_1$. 
\end{theorem} 
\begin{proof}
 Since $\mathcal{I}$ is a compact and connected space, it suffices to prove that if $H(x, t) : T \longrightarrow S^{n-1}$ is a continuous map for sufficiently small $t$, then $v_{0} \stackrel {\alpha}\sim v_t$, where $\alpha \in SL_n(C(T))$ and $\alpha$ can be connected to the identity matrix.\\
We choose small $t$ so that $v_{0}(p) + v_{t}(p) \ne 0$ for every $p \in T$. Define $W : T \longrightarrow S^{n-1}$ by $W(p) = \dfrac{v_{0}(p) + v_{t}(p)}{1 + v_{0}(p)v_{t}(p)}$. Since $v_{0}(p)$ and $v_{t}(p)$ are not antipodal, $v_{0}(p)v_{t}(p) \ne -1$. Since any continuous map $T \longrightarrow S^{n-1}$ gives rise to a unimodular row over $C(T)$, we have $v_{0}(p)W(p) = \dfrac{v_{0}(p)v_{0}(p) + v_{0}(p) v_{t}(p)}{1 + v_{0}(p)v_{t}(p)} = \dfrac{1 + v_{0}(p) v_{t}(p)}{1 + v_{0}(p)v_{t}(p)} = 1$. Similarly $v_{t}(p)$ $.W(p) = 1$. By Vaserstein's Lemma, there exists $\alpha \in SL_n(C(T))$ such that $\alpha$ can be connected to the identity matrix and $v_0 \stackrel {\alpha}\sim v_t$.
\end{proof}
\noindent \textbf{Note}:Consider quaternion algebra\\ $Q = \{x_1 + ix_2 + jx_3 + kx_4 ~|~ x_1, x_2, x_3, x_4 \in \mathbb{R}\}$ over $\mathbb{R}$. Let
$q_1 = x_1 + ix_2 + jx_3 + kx_4$. Define $\varphi : Q \longrightarrow Q$ by $T(q) = q_1q$. Then $\varphi$ is a linear transformation. Also $\varphi(1) = x_1 + ix_2 + jx_3 + kx_4,~ \varphi(i) = ix_1 - x_2 - kx_3 + jx_4,~ \varphi(j) = jx_1 + kx_2 - x_3 - ix_4$ and $\varphi(k) = kx_1 - jx_2 + ix_3 - x_4$. Therefore 
\begin{align*}
matrix (\varphi) = \left(\begin{array}{cccc} x_1 & -x_2 & -x_3 & -x_4\\ x_2 & x_1 & -x_4 & x_3\\ x_3 & x_4 & x_1 & -x_2 \\x_4 & -x_3 & x_2 & x_1\end{array}\right).
\end{align*}  In particular, if $q_1 = ix_2 + jx_3 + kx_4$, then 
\begin{align*}
matrix(\varphi) = \left(\begin{array}{cccc} 0 & -x_2 & -x_3 & -x_4\\ x_2 & 0 & -x_4 & x_3\\ x_3 & x_4 & 0 & -x_2 \\x_4 & -x_3 & x_2 & 0\end{array}\right),
\end{align*} which is a skew symmetric matrix. Hence we have a map $\phi' : \mathbb{R}^{3} \longrightarrow S'$, where $S'$ is the set of all $4 \times 4$ skew symmetric matrices over $\mathbb{R}$ defined by
\begin{align*}
\phi'(x, y , z) = \left(\begin{array}{cccc} 0 & -x & -y & -z\\ x & 0 & -z & y\\ y & z & 0 & -x \\z & -y & x & 0\end{array}\right).
\end{align*}  Also we have a map $\phi = \phi'_{| S^2} : S^2 \longrightarrow S$,
where $S$ is the set of all $4 \times 4$ skew symmetric matrices of determinant $1$ defined 
by 
\begin{align*}
\phi(x, y , z) = \left(\begin{array}{cccc} 0 & -x & -y & -z\\ x & 0 & -z & y\\ y & z & 0 & -x \\z & -y & x & 0\end{array}\right).
\end{align*}

Consider $A = \mathbb{R}[X_1, X_2, \cdots, X_n]/I$, where $I$ is an ideal of real algebraic variety in $\mathbb{R}[X_1, X_2$, $\cdots$, $X_n]$.           
 Suppose $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$ are two unimodular rows over $A$ such that $(a_1, a_2, a_3) \stackrel {E_3(A)}\sim (b_1, b_2, b_3)$. Then the corresponding maps $G_{\vec{a}} : V(\mathbb{R}) \longrightarrow S^{2}$ and $G_{\vec{b}} : V(\mathbb{R}) \longrightarrow S^{2}$ are homotopic. Therefore maps $V(\mathbb{R}) \overset{G_{\vec{a}}} \longrightarrow  S^{2} \overset{\phi} \longrightarrow Q$ and $V(\mathbb{R}) \overset{G_{\vec{b}}} \longrightarrow  S^{2} \overset{\phi} \longrightarrow Q$ are also homotopic.

Let $A$ be a commutative ring with identity. Suppose $\vec{a} = (a_1, a_2, a_3)$ is a unimodular row, then there exists $\vec{b} = (b_1, b_2, b_3) \in A^3$ such that $\langle \vec{a}, \vec{b} \rangle = \vec{a}\vec{b}^{t} = \sum_{i = 1}^{3} a_ib_i = 1$. Therefore we have a skew symmetric matrix 
\begin{align*}
V(\vec{a}, \vec{b}) = \left(\begin{array}{cccc} 0 & a_1 & a_2 & a_3\\ -a_1 & 0 & b_3 & -b_2\\ -a_2 & -b_3 & 0 & b_1 \\-a_3 & b_2 & -b_1 & 0\end{array}\right)
\end{align*}.

Let $S =$ Set of all $4 \times 4$ skew symmetric matrices over $A$. Define a relation $\sim$  on $S$ as $\alpha_1 \sim \alpha_2$ if $\alpha_2 = \beta^{t}\alpha_1\beta$ for some $\beta \in E_4(A)$.
Clearly this is an equivalence relation on $S$. On the other hand  $\stackrel {E_3(A)}\sim$ is an equivalence relation on $Um_3(A)$. 

We define a relation $\Psi : Um_3(A)/E_3(A) \longrightarrow S/\sim$ by $\Psi(\vec{a}) = V(\vec{a}, \vec{b})$.

\textbf{Claim}: $\Psi$ is a well defined map.

Let $\vec{a} = (a_1, a_2, a_3)$ and $\vec{a'} = (a'_1, a'_2, a'_3)$ be such that $\vec{a'} = \vec{a}\tau$, for some $\tau \in E_3(A)$

Suppose $\vec{a}\vec{b}^t = \sum_{i = 1}^{3} a_ib_i = 1$. Then 
\begin{align*}
V(\vec{a}, \vec{b}) = \left(\begin{array}{cccc} 0 & a_1 & a_2 & a_3\\ -a_1 & 0 & b_3 & -b_2\\ -a_2 & -b_3 & 0 & b_1 \\-a_3 & b_2 & -b_1 & 0\end{array}\right).
\end{align*}
Since $\vec{a}\vec{b}^t = 1, ~\vec{a}\tau{\tau}^{-1}\vec{b}^t = 1$. This implies that $\vec{a'}\vec{b'}^t = 1$, where $\vec{b'} = \vec{b}({\tau}^{-1})^t$. We know that $\tau = \prod_{i = 1}^{r} E_{ij}(\lambda)$, where $E_{ij}(\lambda), i\ne j$ is the $3 \times 3$ matrix in $SL_3(A)$ which has $1$ as its diagonal entries and $\lambda$
as its $(i, j)^{th}$ entry. So it suffices to prove the claim when 
\begin{align*}
\tau = \left(\begin{array}{cccc} 1 & \lambda & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right).
\end{align*} Then 
\begin{align*}
\tau^{-1} = \left(\begin{array}{cccc} 1 & -\lambda & 0\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right).
\end{align*} Therefore 
\begin{align*}
V(\vec{a'}, \vec{b'}) = \left(\begin{array}{cccc} 0 & a_1 & a_1\lambda + a_2 & a_3\\ -a_1 & 0 & b_3 & -b_2\\ -a_1\lambda-a_2 & -b_3 & 0 & b_1-\lambda b_2 \\-a_3 & b_2 & -b_1 + \lambda b_2 & 0\end{array}\right).
\end{align*} Let 
\begin{align*}
\beta = \left(\begin{array}{cccc} 1 & 0\\ 0 & \tau^{t}\end{array}\right).
\end{align*} Then 
\begin{align*}
\beta^{t}V(\vec{a}, \vec{b}) = \left(\begin{array}{cccc} 0 & a_1 & a_2 & a_3\\ -a_1 & 0 & b_3 & -b_2\\ -a_1\lambda-a_2 & -b_3 & b_3\lambda & a_1-\lambda b_2 \\-a_3 & b_2 & -b_1 & 0\end{array}\right).
\end{align*} Hence 
\begin{align*}
\beta^{t}V(\vec{a}, \vec{b})\beta =  \left(\begin{array}{cccc} 0 & a_1 & a_1\lambda + a_2 & a_3\\ -a_1 & 0 & b_3 & -b_2\\ -a_1\lambda-a_2 & -b_3 & 0 & b_1-\lambda b_2 \\-a_3 & b_2 & -b_1 + \lambda b_2 & 0\end{array}\right) = V(\vec{a'}, \vec{b'}).
\end{align*} Thus $\Psi$ is well defined.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5


%\section{Obstruction of unimodular row to proper Ideal}
%A vector bundle is a topological construction that makes precise the idea of a family of vector spaces parametrized by another space $X$: to every point $x$ of the space $X$ we associate a vector space $V(x)$ in such a way that these vector spaces fit together to from another space of the same kind as $X$, which is then called a vector bundle over $X$. 
%\begin{exm}
%If there is a fixed vector space $V$ such that $V(x) = V$ for all $x \in V$. Then $X \times V$ is a vector bundle over $X$. Such vector bundles are said to be trivial. If $dim(V) = 1$, then $X \times V$ is called line bundle.
%\end{exm}
%\begin{definition}
%A real vector bundle consists of 
%\begin{enumerate}
%\item topological spaces $X$ and $E$
%\item a continuous surjections $\pi : E \longrightarrow X$ (bundle projection)
%\item for every $x \in X$, the structure of a finite dimensional real vector space on the fiber $\pi^{-1}(\{x\})$.  
%\end{enumerate}
%where the following compatibility condition is satisfied:-\\
%for every point in $X$, there is an open neighborhood $U$, a natural number $k$ and a homeomorphism $\phi : U \times \mathbb{R}^{k} \longrightarrow \pi^{-1}(U)$ such that for all $x \in U$
%\begin{itemize}
%\item $(\pi o \phi)(x, v) = x$ for all $v \in \mathbb{R}^{k}$
%\item the map $v \longrightarrow \phi(x, v)$ is an isomorphism between the vector spaces $\mathbb{R}^{k}$ and $\pi^{-1}(\{x\})$. 
%\end{itemize}
%\begin{exm}
%The Cartesian product $X \times \mathbb{R}^{k}$ equipped with the projection $X \times \mathbb{R}^{k} \longrightarrow X$ is called the \textbf{trivial bundle} of rank $k$ over $X$.
%\end{exm}
%\begin{definition}
%Given a vector bundle $\pi : E \longrightarrow X$ and an open subset $U$ of $X$, a continuous function $s : U \longrightarrow E$ such that $(\pi o s)(u) = u$ for all $u \in U$, is called a \textbf{section} of $\pi$ on $U$.\\
%The function $s$ that maps every element $x \in U$ to the zero element of the vector space $\pi^{-1}(\{x\})$.
%\end{definition}
%Essentially a section assign to every point of $U$ with a vector from the attached vector space in a continuous manner.
%
%Let $F(U)$ be the set of all sections on $U$. With the point-wise addition and scalar multiplication Of sections, $F(U)$ becomes itself a real vector space.The collection of these vector spaces is a \textbf{sheaf} of vector spaces on $X$. 
%
%If $s \in F(U)$ and $\alpha : U \longrightarrow \mathbb{R}$ is a continuous map, then $\alpha s$ (point-wise multiplication) is in $F(U)$. Thus $F(U)$ is a module over the ring of continuous real valued function on $U$.\\
%If $\pi^{-1}(U) = U \times \mathbb{R}^{k}$, then sections over $U$ is free i.e. $F(U) \cong C(U, \mathbb{R}^{k}) =  C(U, \mathbb{R})^{k}$. 
%
%More generally a projective module of rank $n$ is a module which is locally free of rank $n$ (\cite{DG}). If $n = 1$ then it is locally free of rank $1$ and a local generator of module is called a locally nowhere vanishing section.
%\begin{exm}
%
%\begin{enumerate}
%\item  Let $P = $ \{Set of all continuous function f on $[0, 1]$ such that \\$f(0) = -f(1)$\} and $A = $ \{Set of all continuous function g on $[0, 1]$ such that \\ $g(0) = g(1)$\}. Then $P$ is a $A$-module with point-wise addition and multiplication.\\
%\textbf{Claim: $P$ is not a free $A$-module but $P$ is locally free.}
%
%To prove claim it is suffices to prove $P \ne \langle f\rangle$.
%Assume contrary i.e. $P = \langle f\rangle$. Since $f \in P,~ f(k) = 0$ for some $0 < k < 1$. Then $(\alpha f)(k) = 0$ for every $\alpha \in A$. Also we can find a continuous function $g \in P$ such that $g(k) \ne 0$. Thus $g \ne \beta f$ for any $\beta \in A$. This shows that $P$ is not a free $A$-module. \\
%Let $U$ be an open set of $[0, 1]$. Then there exists $g \in P$ such that $g(x) \ne o$ for every $x \in U$. Then for any $f \in P$, we have $f = \dfrac{f}{g}g = \lambda g$, where $\lambda \in A$. Thus $P$ is locally free.
%\item Let $A = \mathbb{R}[X]$ and $a \in \mathbb{R}$. Consider $I = \{f \in A ~|~ f(a) = 0$. Then $I$ is an ideal of $A$. Since $A$ is a PID, $I = \langle X-a \rangle$. Consider $I$ as an $A$-module. Then $I$ is locally free on any open set $U$ of $\mathbb{R}$ not containing $a$ and any polynomial which does not vanish on $U$ generates the module of sections.
%
%Since any polynomial $f$ such that $f(a) = 0$ is divisible by $X - a$ i.e. $f(X) = h(X)(X-a)$. Thus $X-a$ generates the module of sections on $\mathbb{R}$. Since we can divide any section by $X - a$, $X - a$ is a nowhere vanishing section on $\mathbb{R}$ of the line bundle corresponding to $I$. This is analogue of L'Hospital Rule.
%
%Suppose $I/I^{2} = \langle \bar{f(X)} \rangle$. Then $f(X) \in \langle X-a \rangle$, so $f(X) = h(X)(X-a)$. Since $f(X) \in I^{2}, ~ h(a) \ne 0$. Thus  $f(X)$ gives a section of a line bundle corresponding to $I$. This section vanishes at those point where $h = 0$. Therefore $f(X) = \langle X-a \rangle \cap h(X)$, where $\langle X-a \rangle + h(X) = A$ and the zeros of the section corresponding to $f$ are the zeros of the polynomial $h$.
%
%More generally, let $A$ be a Noetherian ring and $I \subset A$ be an ideal . Suppose $I = \langle f_1, f_2, \cdots. f_n \rangle + I^{2}$. Then $\langle f_1, f_2, \cdots, f_n \rangle = I \cap I'$, where $I + I' = A$ and we say that section corresponding to $f_1, f_2, \cdots, f_n$ vanishes at $V(I')$. 
%
%In particular if $I = \langle f_1, f_2, \cdots, f_n \rangle$, we say that the section corresponding to $f_1, f_2, \cdots, f_n$ is nowhere vanishing. Also if $I = \langle f_1, f_2, \cdots, f_n \rangle + I^{2}$, then there exists $g \in I^{2}$ such that $I = \langle f_1, f_2, \cdots, f_n, g \rangle$. So $\bar{I} = \langle \bar{f_1}, \bar{f_2}, \cdots, \bar{f_n} \rangle$. Thus the section corresponding to $f_1, f_2, \cdots, f_n$ is nowhere vanishing on $V(c)$. So we are interested  in extending a nowhere vanishing section on $V(c)$ to the whole of $Spec(A)$. Thus we do by adding multiple of $c$ to $f_i$ to obtain $\langle f_1+\lambda_1 c, f_2+\lambda_2 c, \cdots, f_n+ \lambda_n c \rangle = I \cap I''$, where $I'' + (c) = A$ and $ht(I'') \geq n$.
%\item Let $A = \mathbb{R}[X, Y, Z]/(X^{2}+Y^{2}+Z^{2}-1)$. From example $3.14$, an element of \\$P^{*} = Hom(P, A)$ are sections of the tangent bundle on $S^2$ i.e. continuous vector field on $S^2$. In other words for any element $(a, b, c) \in S^2$, we have a vector $(h_1(a, b, c), h_2(a, b, c), h_3(a, b, c))$ corresponding to $f \in P^*$ which is perpendicular to $(a, b, c)$ i.e. an element of tangent plane.
%\end{enumerate}
%\end{exm}
% Consider $A = \mathbb{R}[X, Y]$ and a maximal ideal $\mathfrak{m} = \langle X, Y \rangle$.
% \begin{enumerate}
% \item Suppose $f(X, Y) = Y - m X +$ higher order term. Then $g(X, Y) = Y - m X$ is a tangent line to $f(X, Y)$ at origin.
% \begin{proof}
% Suppose $Y = m'X$ is a tangent line at $(0, 0)$. Put $Y = m'X$ in $f(X, Y)$. We have $f(X, m'X) = (m' - m)X +$ higher order of $X$. This shows that $X = 0$ is a solution of $f(X, m'X)$. Thus $m = m'$. Hence $g(X, Y) = Y - m X$ is a tangent line to $f(X, Y)$ at origin.
% \end{proof}
% \item Let $f(X, Y) \in A$ such that $f(0, 0) = 0$ i. e $f(X, Y) = aX + bY + \cdots$. Then $aX + bY$ is a tangent line to $f$ at $(0, 0)$. This shows that $\bar{f} \in \mathfrak{m}/\mathfrak{m}^{2}$ gives tangent line to $f$ at $(0, 0)$.
% \item $\mathfrak{m}/\mathfrak{m}^{2}$ is a two dimensional vector space generated by $X + \mathfrak{m}^{2}~\&~ Y + \mathfrak{m}^{2}$ over $\mathbb{R}$.
% \item Let $f, g \in \mathfrak{m}$. Then $\bar{f}, \bar{g}$ generate $\mathfrak{m}/\mathfrak{m}^{2}$ if and only if tangent lines to $f$ and $g$ at origin are different.
% \begin{proof}
% Assume $f, g$ generate $\mathfrak{m}/\mathfrak{m}^{2}$. Suppose tangent lines to $f$ and $g$ at origin are same i.e. $\bar{f} = aX + bY + \mathfrak{m}^{2}$ and $\bar{g} = aX + bY + \mathfrak{m}^{2}$(say). By assumption $X + \mathfrak{m}^{2} = a_1 (aX + bY + \mathfrak{m}^{2}) + b_1 (aX + bY + \mathfrak{m}^{2})$. So $X + \mathfrak{m}^{2} = (a_1 + b_1) (aX + bY + \mathfrak{m}^{2}) \Longrightarrow (1 - (a_1 + b_1)a)X +(a_1 + b_1)bY \in \mathfrak{m}^{2} \Longrightarrow 1 - (a_1 + b_1)a = 0 ~\&~ (a_1 + b_1)b = 0$. If $a_1 + b_1 = 0$, then $1 = 0$ which is absurd. If $b = 0$, then $Y + \mathfrak{m}^{2} = a_2 (aX + \mathfrak{m}^{2}) + b_2 (aX + \mathfrak{m}^{2}) \Longrightarrow Y - (a_2 + b_2)aX \in \mathfrak{m}^{2}$ which is not possible. Hence tangent lines to $f$ and $g$ at origin are different. \\
% Conversely suppose tangent lines to $f$ and $g$ at origin are different. Then we have $f + \mathfrak{m}^{2} = a_1 X + b_1 Y + \mathfrak{m}^{2}$ and $g +\mathfrak{m}^{2} = a_2 X + b_2 Y + \mathfrak{m}^{2}$. This implies that $X + \mathfrak{m}^{2} = (\dfrac{b_2}{b_2a_1 - a_2 b_1})(f + \mathfrak{m}^{2}) + (\dfrac{b_1}{b_2a_1 - a_2 b_1})(g + \mathfrak{m}^{2}) ~\&~ Y + \mathfrak{m}^{2} = (\dfrac{a_2}{b_1a_2 - a_1 b_2})(f + \mathfrak{m}^{2}) + (\dfrac{a_1}{b_1a_2 - a_1 b_2})(g + \mathfrak{m}^{2})$. Thus $\bar{f}, \bar{g}$ generate $\mathfrak{m}/\mathfrak{m}^{2}$.  
% \end{proof}
% Suppose $\mathfrak{m}/\mathfrak{m}^{2} = \langle f, g \rangle$, where $f = X mod \mathfrak{m}^{2}$ and $g = Y mod \mathfrak{m}^{2}$. Thus $f = (1 + p)X + qY$ and $g = p'X + (1 + q')Y$, where $p, p', q, q' \in \langle X, Y \rangle$. Take $\alpha = \left(\begin{array}{cccc} 1+p & q\\ p' & 1+q'\end{array}\right)$. Thus $\alpha \left(\begin{array}{cccc} X\\ Y\end{array}\right) = \left(\begin{array}{cccc} f\\ g\end{array}\right)$ and $det(\alpha) = 1-h$, where $h \in \langle X, Y \rangle$. Since $\langle f, g \rangle \subset \langle X, Y \rangle, {\langle f, g \rangle}_{1-h} \subset \langle X, Y \rangle_{1-h}$. On other hand, we have \\$\left(\begin{array}{cccc} X\\ Y\end{array}\right) = \dfrac{adj(\alpha)}{1-h}\left(\begin{array}{cccc} f\\ g\end{array}\right)$. Therefore $\langle X, Y \rangle_{1-h} \subset {\langle f, g \rangle}_{1-h}$. Hence \\$\langle X, Y \rangle_{1-h} = {\langle f, g \rangle}_{1-h}$. This shows that  
% ideals $\langle X, Y \rangle$ and ${\langle f, g \rangle}$ are same where $(1-h)$ does not vanish.
% \\ \textbf{Claim: $\langle X, Y \rangle = \langle f, g, h \rangle$}
% \begin{proof}
% Since $\langle f, g \rangle \subset \langle X, Y \rangle ~\&~ h \in \langle X, Y \rangle,~ \langle f, g, h \rangle \subset \langle X, Y \rangle$. Also we have $\alpha \left(\begin{array}{cccc} X\\ Y\end{array}\right) = \left(\begin{array}{cccc} f\\ g\end{array}\right) \Longrightarrow (1-h) \left(\begin{array}{cccc} X\\ Y\end{array}\right) = adj(\alpha)\left(\begin{array}{cccc} f\\ g\end{array}\right)$ i.e. $\langle X, Y \rangle (1-h) \subset \langle f, g \rangle \Longrightarrow \langle X, Y \rangle (1-h) \subset \langle f, g, h \rangle$. Since $\langle X, Y \rangle (h) \subset \langle f, g, h \rangle$. So $\langle X, Y \rangle (1-h + h) \subset \langle f, g, h \rangle \Longrightarrow \langle X, Y \rangle  \subset \langle f, g, h \rangle$. Hence the claim. \\ Since $h \in \langle X, Y \rangle$ and $(1-h) \left(\begin{array}{cccc} X\\ Y\end{array}\right) = adj(\alpha)\left(\begin{array}{cccc} f\\ g\end{array}\right),~ h(1-h) \in \langle f, g \rangle$
% \end{proof}
%The general form of previous discussion is following:\\
%Let $A$ be a Noetherian ring and $I \subset A$ be an ideal. Suppose $I = \langle a_1, a_2, \cdots, a_n \rangle + I^2$. Then $I = \langle a_1, a_2, \cdots, a_n, e \rangle$, where $e(1-e) \in \langle a_1, a_2, \cdots, a_n \rangle$.\\
%This shows that `` If $I/I^{2}$ is generated by $n$ elements then $I$ is generated by $n+1$ elements. So there is a question that when $I$ is generated by $n$ elements. Now we give obstruction of some results over unimodular rows (to see proof of following, we will refer \cite{RR})
%\begin{itemize}
%\item Let $A$ be a Noetherian ring of dimension $d$. Suppose $\langle a_1, a_2, \cdots, a_n \rangle = A$, where $n \geq d+2$. Then there exist $c_1, c_2, \cdots, c_{n-1}$ such that $\langle a_1 + c_1 a_n, a_2 + c_2 a_n, \cdots, a_{n-1}+ c_{n-1} a_n \rangle = A$. In other words, $(a_1, a_2, \cdots, a_n)$ is elementary completable.\\ The obstruction of this results over $I/I^{2}$ is :\\Let $A$ be a Noetherian ring, $J \subset A$ an ideal. Suppose $J = \langle a_1, a_2, \cdots, a_n, s \rangle$, where $s \in J^{2}$. then there exist $c_1, c_2, \cdots, c_{n-1}$ such that $\langle a_1 + c_1 a_n, a_2 + c_2 a_n, \cdots, a_{n-1}+ c_{n-1} a_n \rangle = J \cap J'$, where $J' + \langle s \rangle = A$ and $ht(J') \geq n$. In particular, if $n \geq dim(A) + 1$ then $J = \langle a_1 + c_1 a_n, a_2 + c_2 a_n, \cdots, a_{n-1}+ c_{n-1} a_n \rangle$.
%\item In a semilocal ring $A$ any unimodular row of length $n \geq 2$ is completable.\\ The obstruction of this results over $I/I^{2}$ is :\\Let $A$ be a Noetherian semilopcal ring and $I$ be an ideal of $A$. Suppose $I/I^{2}$ is generated by $n$ elements. Then $I$ is generated by $n$ elements.
%\item Let $A$ be a Noetherian ring and $(a_1, a_2, \cdots, a_n) \in Um_{n}(A)$. Suppose $n \geq dim(A/Jac(A)) + 2$. Then $(a_1, a_2, \cdots, a_n)$ is elementary completable.\\ The obstruction of this results over $I/I^{2}$ is :\\Let $A$ be a Noetherian ring and $I$ be an ideal of $A$. Suppose $I/I^{2}$ is generated by $n$ elements, where $n \geq dim(A/Jac(A)) + 2$. Then $I$ is generated by $n$ elements.
%\item Let $A$ be a local ring and $(f_1(X), f_2(X), \cdots, f_n(X))$ be a unimodular row in $A[X]$ with one entry monic. Then $(f_1(X), f_2(X), \cdots, f_n(X))$ is completable.\\ The obstruction of this results over $I/I^{2}$ is :\\Let $A$ be a Noetherian local ring and $I$ be an ideal of $A[X]$. Suppose $I/I^{2}$ is generated by $n$ elements. Then $I$ is generated by $n$ elements.
%\item Let $A$ be a Noetherian domain of dimension $d$ and $(f_1(X), f_2(X), \cdots, f_n(X))$ be a unimodular row in $A[X]$, where $n \geq d+2$. Then $(f_1(X), f_2(X), \cdots, f_n(X))$ is completable.\\ The obstruction of this results over $I/I^{2}$ is :\\ Let $A$ be a Noetherian domain of dimension $d$ and $I$ be an ideal of $A[X]$. Suppose $I/I^{2}$ is generated by $n$ elements, where $n \geq d+1$. Then $I$ is generated by $n$ elements.
%
%
%\end{itemize} 
%
% 
%   
% \end{enumerate}

 \begin{thebibliography}{XXX}
%\bibitem[2]{LA}
%L. N. Vaserstein and A. A. Suslin ;
%\newblock{Serre's problem on projective modules over polynomial rings and algebraic K - theory},
%\newblock{Math. USSR Izvestija 10 (1976), 937 - 1001}.
\bibitem[1] {HB}
Bass H.; 
\newblock{K-theory and stable algebra},
\newblock{Publications Mathematiques. Institut de Hautes
Etudes Scientifiques, 22 (1964), 5 - 60}.
\bibitem[2]{RR}
Basu R, Sridharan R; 
\newblock{On Forster's conjecture and related results},
\newblock{Punjab Univ. Res. J. (Sci.), { 57}, 2007, 13 - 66}.
\bibitem[3]{SD}
Deo S.;
\newblock {Algebraic topology (A primer)},
\newblock {Hindustan Book Agency, New Delhi, 2003}.
\bibitem[4]{MK}
Krusemeyer M;
\newblock{Skewly completable rows and a theorem of Swan Towber},
\newblock{Communication in Algebra, 4 (7) (1975), 657 - 663}.
\bibitem[5]{TL}
Lam T Y;
\newblock{Serre's Conjecture},
\newblock{Lecture Notes in Mathematics, 635, Springer - Verlag Berlin Heidelberg, New York, 1978}.
\bibitem[6]{JR}
Munkres J R;
\newblock{Topology, A first course},
\newblock{Prentice Hall of India, Private Limited (1996)}.
%\bibitem[6]{DG}
%Northcott D. G.;
%\newblock{Finite free resolutions},
%\newblock{Camb. Univ. Press, 1976}.
\bibitem[7]{RS}
Simha R R;
\newblock {Spheres and orthogonal groups},
\newblock {Bombay Mathematical Colloquium, Bulletin 14, No. 1, January 1998}.
\bibitem[8]{VS}
Vaserstein L N, Suslin A A;
\newblock {Serre's problem on projective module over polynomial ring and algebraic K - theory},
\newblock {Izv. Akad. Nauk SSSR, ser. mat. 40 (1976) 993-1054}.
\bibitem[9]{HY}
Young G S, Hocking J G;
\newblock {Topology},
\newblock {Dover Publication, Inc. Newyork, 1961}.
 \end{thebibliography}






%\end{definition} 

\end{document}
1506.02356.extracted_bib
[2]{LA %L. N. Vaserstein and A. A. Suslin ; %\newblock{Serre's problem on projective modules over polynomial rings and algebraic K - theory, %\newblock{Math. USSR Izvestija 10 (1976), 937 - 1001.
1506.02356.extracted_bib
[1] {HB Bass H.; \newblock{K-theory and stable algebra, \newblock{Publications Mathematiques. Institut de Hautes Etudes Scientifiques, 22 (1964), 5 - 60.
1506.02356.extracted_bib
[2]{RR Basu R, Sridharan R; \newblock{On Forster's conjecture and related results, \newblock{Punjab Univ. Res. J. (Sci.), { 57, 2007, 13 - 66.
1506.02356.extracted_bib
[3]{SD Deo S.; \newblock {Algebraic topology (A primer), \newblock {Hindustan Book Agency, New Delhi, 2003.
1506.02356.extracted_bib
[4]{MK Krusemeyer M; \newblock{Skewly completable rows and a theorem of Swan Towber, \newblock{Communication in Algebra, 4 (7) (1975), 657 - 663.
1506.02356.extracted_bib
[5]{TL Lam T Y; \newblock{Serre's Conjecture, \newblock{Lecture Notes in Mathematics, 635, Springer - Verlag Berlin Heidelberg, New York, 1978.
1506.02356.extracted_bib
[6]{JR Munkres J R; \newblock{Topology, A first course, \newblock{Prentice Hall of India, Private Limited (1996). %
1506.02356.extracted_bib
[6]{DG %Northcott D. G.; %\newblock{Finite free resolutions, %\newblock{Camb. Univ. Press, 1976.
1506.02356.extracted_bib
[7]{RS Simha R R; \newblock {Spheres and orthogonal groups, \newblock {Bombay Mathematical Colloquium, Bulletin 14, No. 1, January 1998.
1506.02356.extracted_bib
[8]{VS Vaserstein L N, Suslin A A; \newblock {Serre's problem on projective module over polynomial ring and algebraic K - theory, \newblock {Izv. Akad. Nauk SSSR, ser. mat. 40 (1976) 993-1054.
1506.02356.extracted_bib
[9]{HY Young G S, Hocking J G; \newblock {Topology, \newblock {Dover Publication, Inc. Newyork, 1961.