Existence of a canonical isomorphism for ($\operatorname{Hom}$-$\otimes$ adjunction).

Question

Here is the question I want to answer:

Let $U,V, W$ be finite dimensional vector spaces over the field $F.$ Prove that there is a canonical isomorphism $$\operatorname{Hom}(U \otimes V, W) = \operatorname{Hom}(U, V^* \otimes W).$$ Deduce from this formula that there is a canonical isomorphism $$\operatorname{Hom}(U \otimes V, W) = \operatorname{Hom}(U, \operatorname{Hom}(V ,W)).$$

My thoughts:

I know from this question $ \Phi : \operatorname{Hom}(U \otimes V,W) \to \operatorname{Hom}(U,\operatorname{Hom}(V,W)) $ Isomorphism the solution of the second part immediately but I want to answer the question in the order it is stated, so could someone tell me how can I show the existence of this canonical isomorphism $$\operatorname{Hom}(U \otimes V, W) = \operatorname{Hom}(U, V^* \otimes W) $$ and then deduce from it the isomorphism $$\operatorname{Hom}(U \otimes V, W) = \operatorname{Hom}(U, \operatorname{Hom}(V ,W)).$$ Could anyone help me please in this(the general steps are not completely clear to me)?

Answer 1

The exercise basically forces you to cheat, since the first isomorphism only works in finite dimensions, while the second not. $\newcommand{\Hom}{\operatorname{Hom}}$

1. First, we have a well-known isomorphism $\phi \colon V^* \otimes W \to \Hom(V,W)$ such that $\phi(f \otimes w) = f(\_)w$. I elaborate a little bit on that: First, $\phi$ is well-defined since the map $V^* \times W \to \Hom(V,W)$ given by $(f,w) \mapsto f(\_)w$ is bilinear. Now, if we pick bases $\{v_i\}_i$ (with $\{f_i\}_i$ as its dual) and $\{w_j\}_j$ for $V$ and $W$, then $\{f_i \otimes w_j\}_{(i,j)}$ is a basis for $V^* \otimes W$, and $\phi(f_i \otimes w_j)$ corresponds precisely to the matrix with zeroes everywhere except a one at the $(j,i)$-entry. Thus, $\phi$ sends a basis to a basis, and hence it is an isomorphism. (Note that this doesn’t works in infinite dimensions.)
2. Since $\phi$ is an isomorphism, then so is $$\phi \circ (\_) \colon \Hom(U,V^* \otimes W) \to \Hom(U,\Hom(V,W)).$$ Thus, once we do the first step of the exercise, the second follows from this, right?
3. The space $\Hom(U,\Hom(V,W))$ is “the same as” the space $\operatorname{Bil}(U,V;W)$ of bilinear maps $U \times V \to W$. How? Well, an element $g$ of the first space induces a bilinear map $\tilde g \colon U \times V \to W$ given by $\tilde g(u,v) = g(u)(v)$, and a bilinear map $b \colon U \times V \to W$ induces a linear map $b ^\sharp \colon U \to \Hom(V,W)$ by $b^\sharp(u) = b(u,\_)$. It is easy to see that the assignments $g \mapsto \tilde g$ and $b \mapsto b^\sharp$ are inverses to each other, and moreover, they are linear.
4. Finally, the universal property of the tensor product says that for each $b \in \operatorname{Bil}(U,V;W)$ there exists a unique $h \in \Hom(U \otimes V,W)$ such that $h \circ \otimes = b$, where $\otimes \colon U \times V \to U \otimes V$ sends a pair $(u,v)$ to $u \otimes v$. In other words, the universal property says that the map $$(\_) \circ \otimes \colon \Hom(U \otimes V,W) \to \operatorname{Bil}(U,V;W)$$ is bijective. Since we can easily prove that the above map is linear, it follows that it is an isomorphism.

Then, as you see, from steps 2, 3 and 4 we have that $$\begin{align*} \Hom(U,V^* \otimes W) &\cong \Hom(U,\Hom(V,W)) \\ &\cong \operatorname{Bil}(U,V;W) \\ &\cong \Hom(U \otimes V,W) \end{align*}$$ but the steps 3 and 4 already gives us $\Hom(U \otimes V,W) \cong \Hom(U,\Hom(V,W))$.

If you don’t want to cheat like me, you can directly prove that $$\begin{align*} \Hom(U,V^* \otimes W) &\longrightarrow \Hom(U \otimes V,W) \\ f &\longmapsto \big( u \otimes v \mapsto \phi(f(u))(v) \big) \end{align*}$$ is an isomorphism (try to write a formula for its inverse) and then conclude from 2 that $\Hom(U \otimes V,W) \cong \Hom(U,\Hom(V,W))$.

Answer 2

Write it out explicity. Here is a sketch:

$1).\ $ if $f\in \operatorname{Hom}(U \otimes V, W)$ then for each $u\in U,\ f_u:= f(u,\cdot):V\to W.$

$2).\ $ there is the adjoint $(f_u)^*:W^*\to V^*: (f_u)^*(w^*)(v)=w^*((f_u)(v))=w^*(f(u,v)).$

$3).\ $ for each $w\in W$ we have the evaluation map $\operatorname {eval}_w:W^*\to F:\operatorname {eval}_w(\varphi)=\varphi(w).$

So it makes sense to define

$4).\ \Phi:\operatorname{Hom}(U \otimes V, W) \to \operatorname{Hom}(U, V^* \otimes W)$ by $\Phi(f)=g$ where $g(u)=((f_u)^*(\operatorname {eval}_{f(u,v)}),f(u,v)).$