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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.)

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$\newcommand{\Hom}{\operatorname{Hom}}$ 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.

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Viktor Vaughn
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