Q: If a Set of Cones is Representable, then there is a Limit I am following Maclane's proof ot the existence of a pointwise right Kan extension, and I am stuck on a probably easy fact: 
Let $F:J\to A$ be a functor, and suppose that, for all $a\in A$, there is an $L\in A$ and an isomorphism $\tag1[a,L]\cong Cone(a,F)$
i.e. the set of cones to $F$ is representable. Then $F$ has a limit in $A$. 
We have a the unique correspondence $f\leftrightarrow \left \{ f_i \right \}_{i\in J}$ where $f:a\to L$ and $f_i:a\to F_i$ for each $i\in J$.
Set $a=L$ and consider the correspondence $1_{L}\leftrightarrow \left \{ \lambda _i \right \}_{i\in J}$. 
To prove that $\left \langle L, \left \{ \lambda _1 \right \}_{i\in J} \right \rangle$ is a limit cone, it is enough to show that $\lambda _i\circ f=f_i$ and this follows easily enough if $(1)$ is natural in $a$, for then we have the commutng square
\begin{matrix}[L,L] &\stackrel{\phi '}{\rightarrow}&[\Delta_L,F]\\\downarrow{}&&\downarrow{}\\ [a,L]& \stackrel{\phi }{\rightarrow}&[\Delta_a,F]\end{matrix}
where $\phi $ and $\phi '$ are the isomorphisms that establish the above correspondences. Following $1_L$ around the square gives the result. 
My question is: what happens if the isomorphisms are not natural in $a$? In MacLane's proof, in which $(1)$ obtained in several steps, the naturality is not clear, and is not explicitly advertised. So I am wondering if it is obvious or unnecessary. 

A: In general, you can conclude very little from a pointwise isomorphism of sets $[a,L]\cong Cone(a,F)$ without naturality.  For instance, suppose $J$ has one object and no non-identity morphisms, so $F$ just picks out an object $b$, a limit of $F$ is an object with an isomorphism to $b$, and $Cone(a,F)\cong [a,b]$.  Then all you know from a non-natural isomorphism is that you have some object $L$ such that $[a,L]$ has the same cardinality as $[a,b]$ for all $a$.  This certainly doesn't imply that $L\cong b$ in general.  For instance, in the category of vector spaces over an infinite field $k$, you can see that $|[a,L]|=|[a,b]|$ for any $a$ whenever $\dim L$ and $\dim b$ are both positive and finite.
