Hello,

I'm looking for an invariant to distinguish the homeomorphism types of homotopy equivalent spaces. Specifically, how does one show that the total spaces of the tangent bundle to $S^2$ and the trivial bundle $S^2 \times R^2$ are not homeomorphic? (I am not asking for a proof that $TS^2$ is not the trivial bundle.)

Also, is there a way to reduce the question, "Are the total spaces of two vector bundles homeomorphic" to "Are the associated sphere bundles homeomorphic"? In the case of $TS^2$ and $S^2\times R^2$ it's not too difficult to show that the sphere bundles are not homeomorphic, and I'm wondering if there's a way to leverage that.

Thanks,

Zygund

This is more or less equivalent to Ryan's comment but with more details and a slightly different point of view.

Let $X$ be the total space of the tangent bundle, and put $Y=S^2\times\mathbb{R}^2$.  If $X$ and $Y$ were homeomorphic, then their one-point compactifications would also be homeomorphic.  We will show that this is impossible by considering their cohomology rings.

Put $X'=\{(p,q)\in S^2\times S^2 : p+q\neq 0\}$.  There is a homeomorphism $f:X\to X'$ given by $f(u,v)=((\|v\|^2-1)u+2v)/(\|v\|^2+1)$ (a variant of stereographic projection).  It follows that $X_\infty$ can be obtained from $S^2\times S^2$ by collapsing out the antidiagonal.  We have $H^*(S^2\times S^2)=\mathbb{Z}[a,b]/(a^2,b^2)$ and it follows that $H^*(X_\infty)$ is the subring generated by $1$, $a+b$ and $ab$.  In particular, the squaring map from $H^2$ to $H^4$ is nonzero.

However, $Y$ can be identified with $(S^2\times S^2)\setminus (S^2\times\{point\})$, so $H^*(Y_\infty)$ is isomorphic to the subring generated by $1$, $a$ and $ab$, so the squaring map $H^2\to H^4$ is zero. 

Note that the tangent bundle plus a rank-one trivial bundle is trivial, so the suspensions of $X_\infty$ and $Y_\infty$ are homeomorphic.
