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In my differential geometry class we've now moved onto algebraic/differential forms and to begin the section we're doing a quick and easy review of dual vector spaces. On a problem sheet I am confronted by the question:

"Let $V$, $W$ be vector spaces over $k$ of dimensions $n$ and $m$ respectively. Show that the vector space of linear maps $V\rightarrow W$ may be identified with the space of linear maps $V\times W^*\rightarrow k$, where $W^*$ is the dual space of $W$."

There's bound to be a down-to-earth explanation of this which I'm sure we'll see in class and for which I'm not interested in seeing an answer on this site. However, I think I've got a categorical proof which I'd like to see checked. It uses that in a Cartesian closed category $\mathbf{C}$ we have $\operatorname{Hom} (A\times B, C)\cong \operatorname{Hom} (A, C^B)$ for objects $A,B,C\in \operatorname{ob}(\mathbf{C})$.

Note that $W^* = \operatorname{Hom} (W,k)$ in $k$-$\mathbf{FDVect}$, the category of finite-dimensional vector spaces over $k$. Then we have

$\operatorname{Hom} (V\times W^*, k)\cong\operatorname{Hom} (V, k^{W^*})$. But $k^{W^*}=\operatorname{Hom}(W^*, k)= W^{**}\cong W$.

Does this work? The problem is that the category of finite-dimensional vector spaces is not Cartesian closed - it apparently doesn't have a good definition of exponential object so I'm not sure if the first isomorphism here is justified.

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The question is incorrect, you need a tensor product in place of the product. With that modification you can use/prove that

$\mathrm{Hom}_\mathrm{FDVect}(V,W) \cong W \otimes V^*$

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You are correct that the category of finite-dimensional vector spaces is not cartesian closed. However, it is closed monoidal: this is like being cartesian closed, but with the cartesian product replaced with an arbitrary monoidal product, here the tensor product. The essential adjunction now takes the form

$$\text{Hom}(U \otimes V, W) \cong \text{Hom}(U, [V, W])$$

where $[V, W]$ denotes the vector space of linear maps $V \to W$. In the particular case of finite-dimensional vector spaces, but not in more general settings, it is furthermore the case that $[V, W] \cong V^{\ast} \otimes W$.

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