Showing $\hom_{F}\left(\prod_{i=1}^{n}V_i\ ,\prod_{j=1}^{m}W_j\right)\cong\prod_{i=1}^{n}\prod_{j=1}^{m}\hom_{F}(V_i,W_j)$.

Question

$\newcommand{\Hom}{\text{Hom}}$ I want to prove the following result for finite products of vector spaces:

$$\Hom_{F}\left(\prod_{i=1}^{n}V_i\ ,\prod_{j=1}^{m}W_j\right)\cong\prod_{i=1}^{n}\prod_{j=1}^{m}\Hom_{F}(V_i,W_j).\tag{1}$$

Here, $A\cong B$ denotes $A$ and $B$ are isomorphic. I am aware of the following two results:

$$\Hom_{F}\left(V,\prod_{j=1}^{m}W_j\right)\cong\prod_{j=1}^{m}\Hom_{F}(V,W_j).\tag{2}$$ $$\Hom_{F}\left(\prod_{i=1}^{n}V_i\ ,W\right)\cong\prod_{i=1}^{n}\Hom_{F}(V_i,W).\tag{3}$$

My Attempt:
Let $\Psi:\Hom_{F}\left(\prod_{i=1}^{n}V_i\ ,\prod_{j=1}^{m}W_j\right)\rightarrow\prod_{i=1}^{n}\prod_{j=1}^{m}\Hom_{F}(V_i,W_j)$ be defined by $\Psi(T)=(\pi_j T\theta_i)_{i=1,\dots,n,\ j=1,\dots,m}$. I wanted to show $\Psi$ is an isomorphism. Here, $\pi_j$ and $\theta_i$ are the $j$-th projection and the $i$-th injection defined as follows:

$\pi_p:\prod_{k\in\Delta} V_k\rightarrow V_p$ such that $\pi(f)=f(p)$ for all $f$ in the product space.
$\theta_q:V_q\rightarrow\prod_{k\in\Delta}V_k$ such that $\theta_q(\alpha)(i)=\begin{cases} \alpha, & \text{if} & i=q\\ 0, & \text{if} & i\ne q \end{cases}$.
Here, $\Delta$ is an arbitrary indexing set (which, in this context, is finite).

I could show that $\Psi$ is injective by showing $\Psi(S)=\Psi(T)$ implies $S=T$ for all $S,T$. But I do not see we can show that it is also surjective. I was trying to set $T$ equal to a sum as is usually done in the proofs of $(2) \& (3)$ but cannot find a suitable form that works.

Some help\alternate approach would be highly appreciated. TIA.

Answer 1

Since I got asked to post it as answer: (sorry for being very handwavy, currently dont have the time to do all the diagrams and dont know if tikz works here)

if you have a tupel of morphisms on the left (btw. thank god that those products are finite, otherwise you get into real trouble) there EXISTS by universal property a UNIQUE map making the diagram commute, that is welldefinednes and injectivity of your map.

Now assume you got a map on the right, then composing with all the projections and inclusions from your diagram (again, the products are finite, so products are coproducts and all is fine) gives you a diagram that would suffice the universal property and BAM there is surjectivity (uniqueness tells you the induced morphism must be the morphism you started with).

Final funfact: the same argument holds in every additive cat!

Answer 2

If you set $V = \prod V_i$ and $W = \prod W_j$, and believe (2) and (3), then $$\DeclareMathOperator{\Hom}{Hom} \Hom(V,W) \stackrel{(2)}{=} \prod_i \Hom(V_i, W) \stackrel{(3)}{=} \prod_i \prod_j \Hom(V_i, W_j).$$ If you follow the given isomorphisms of (2) and (3), this should hopefully also give you the isomorphism you require.

Here is a subtle note: To only prove that the spaces are isomorphic as $F$-vector spaces, it would actually be sufficient to count their dimensions. Suppose $\dim V_i = n_i, \dim W_j = m_j$, then both sides have dimension $$\left(\sum_i n_i\right)\left(\sum_j m_j\right) = \sum_{i,j} n_i m_j,$$ because dimensions are additive for (finite) products and multiplicative when applying $\Hom$ (think of matrices). So at this point is just follows from linear algebra, that the two spaces are abstractly isomorphic. However, this is probably not the result you want to have, because you want a special homomorphism to be an isomorphism, which has certain naturality properties.