It has just occurred to me that most of my intuition for tensor products is derived from the special case of finite-dimensional vector spaces, so I'm wondering which properties I've taken for granted are true in general, and which are not.
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Let $U$ and $V$ be $k$-vector spaces, possibly infinite-dimensional. Does it remain true that $U^* \otimes V \cong \textrm{Hom}(U, V)$ naturally in $U$ and $V$?
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Let $A, B, C$ be objects in an abelian category, or better, a monoidal closed category. Is it true that $\textrm{Hom}(A, B \otimes C) \cong \textrm{Hom}(A, B) \otimes \textrm{Hom}(A, C)$ naturally in $A, B, C$? (Motivation: $\textrm{Hom}(A, -)$ preserves (cartesian) products.)
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In the same context as above, is there a bifunctor $\mathscr{F}(-, -)$ such that $\textrm{Hom}(A, C) \otimes \textrm{Hom}(B, C) \cong \textrm{Hom}(\mathscr{F}(A, B), C)$ naturally in $A, B, C$? (Motivation: $\textrm{Hom}(-, C)$ maps coproducts to products.)
Best Answer
No. As Theo says in the comments, the elements of $U^{\ast} \otimes V$ are precisely the finite-rank maps in $\text{Hom}(U, V)$.
This isn't even true in finite dimensions. The dimension of the LHS grows linearly in $\dim A$ but the dimension of the RHS grows quadratically in $\dim A$.
This also isn't even true in finite dimensions. The dimension of the LHS grows quadratically in $\dim C$ but the dimension of the RHS grows linearly in $\dim C$.
You seem to be under the mistaken impression that the tensor product is supposed to behave like a product. It isn't. Abstractly it comes from the tensor-hom adjunction
$$\text{Hom}(A \otimes B, C) \cong \text{Hom}(A, \text{Hom}(B, C))$$
and concretely it comes from wanting the free vector space functor $\text{Set} \to \text{Vect}$ to be (lax?) monoidal.
Working with naked infinite-dimensional vector spaces is asking for trouble. See topological tensor product for appropriate substitutes for topological vector spaces.