Isomorphism of some Hom spaces Let $U,V,W$ be vector spaces over same field $F$. Let $Hom(-,-)$ denote set of $F$-linear maps from domain to co-domain. I was thinking on following isomorphism if it is true! 
$${\rm Hom}(U, {\rm Hom}(V,W)\cong {\rm Hom}({\rm Hom}(U,V), W)$$
If all vector spaces are finite dimensional then the isomorphism holds just by comparison of dimension. But, here I want to know whether there is a natural isomorphism above. Also, if we replace vector spaces by modules over same commutative ring with $1$ then still this isomorphism is valid?
(I am beginning undergraduate, and I have no ideas of above type of problems.)
 A: $\newcommand{\Hom}{\operatorname{Hom}}$
I doubt the isomorphism is natural even for finite-dimensional vector spaces.  By tensor-Hom adjunction, we have
$$
\Hom(U, \Hom(V,W))\cong \Hom(U \otimes V, W) \, .
$$
Now if $U$ and $V$ are finite-dimensional, then $U \otimes V \cong \Hom(U^*,V)$ (see this question), so
$$
\Hom(U, \Hom(V,W))\cong \Hom(U \otimes V, W) \cong \Hom(\Hom(U^*,V),W) \, .
$$
You are asking if this is isomorphic to $\Hom(\Hom(U,V),W)$, which would be true for instance if $U^* \cong U$.  This is true for finite-dimensional vector spaces, but the isomorphism $U^* \cong U$ is not natural.  If $U$ is an inner product space (or, more generally, is equipped with a nondegenerate bilinear form), then $U \cong U^*$ naturally by the map $u \mapsto \langle \cdot, u \rangle$.
If $U$ is finite-dimensional and we take $V = W = F$, then
\begin{align*}
\Hom(\Hom(U,V),W) &= \Hom(\Hom(U,F),F) = \Hom(U^*,F) = U^{**} \cong U\\
\Hom(\Hom(U^*,V),W) &= \Hom(\Hom(U^*,F),F) = \Hom(U^{**},F) = U^{***} \cong U^*
\end{align*}
so I guess the isomorphism can't be natural in general.
