Axler: Prove $\mathcal{L}(V_1\times\ldots\times V_m,W)$ and $\mathcal{L}(V_1,W)\times\ldots\times \mathcal{L}(V_m,W)$ are isomorphic vector spaces.

Question

Suppose $V_1,...,V_m$ are vector spaces. Prove that $\mathcal{L}(V_1\times\ldots\times V_m,W)$ and $\mathcal{L}(V_1,W)\times\ldots\times \mathcal{L}(V_m,W)$ are isomorphic vector spaces.

There are other questions (here and here) about this problem but they don't contain full proofs with all the intermediate details. Answers to those questions present hints or high-level discussions about approaches to these types of problems.

Initially, I thought all I had to do to solve this problem was show that the dimensions of the two vector spaces are the same. This is very easy to do.

There are two issues with this, however.

1. The problem does not assume $V_1,...,V_m$ are finite-dimensional.

2. After reading this answer I was convinced that it is important to actually find an isomorphism.

Here is my attempt at solving this problem, which I found very tricky.

Let $\alpha_{i,1},\alpha_{i,2},\ldots,\alpha_{i,\infty}$ be a basis for $\mathcal{L}(V_i,W)$.

Define the linear map $\Gamma: \mathcal{L}(V_1,W)\times\ldots\times\mathcal{L}(V_m,W)\to \mathcal{L}(V_1\times\ldots\times V_m,W)$ by

$$\Gamma(f_1,\ldots,f_m)=\sum\limits_{i=1}^m f_i$$

Let's prove that this $\Gamma$ is a bijection.

First we show that it is injective.

Suppose

$$\Gamma(f_1,\ldots,f_m)=\Gamma(h_1,\ldots,h_m)\tag{1}$$

Then, $\sum\limits_{i=1}^m f_i=\sum\limits_{i=1}^m h_i$.

Now, since $f_i\in\mathcal{L}(V_i,W)$ and $h_i\in\mathcal{L}(V_i,W)$ then

$$f_i=\sum\limits_{j=1}^\infty a_{ij}\alpha_{ij}$$

$$h_i=\sum\limits_{j=1}^\infty b_{ij}\alpha_{ij}$$

Then, from (1) we have

$$\sum\limits_{i=1}^m\sum\limits_{j=1}^\infty (a_{ij}-b_{ij})\alpha_{ij}=0$$

The $m$ inner sums represent $m$ vectors.

If any of them are non-zero, then they are linearly independent and the outer sum is non-zero since it is a linear combination of such vectors with non-zero coefficients. This case is ruled out.

The inner vectors must thus be zero vectors.

Thus,

$$\sum\limits_{j=1}^\infty (a_{ij}-b_{ij})\alpha_{ij}=0$$

for $i=1,\ldots, m$ and so

$$a_{ij}=b_{ij}$$

for $i=1,\ldots,m$ and $j=1,\ldots,\infty$.

Thus, $f_i=h_i$ and $\Gamma$ is injective.

Next, we show that $\Gamma$ is surjective.

Let $g\in\mathcal{L}(V_1\times\ldots\times V_m,W)$.

Let $(v_1,\ldots,v_m)\in V_1\times\ldots\times V_m$ and

$$g(\alpha_{1j},0,\dots,0)=w_{1j}$$ $$\ldots$$ $$g(0,0,\dots,0,\alpha_{mj})=w_{mj}$$

for $j=1,\dots,\infty$.

Let $(v_1,\ldots,v_m)=\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j},\ldots \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )\in(V_1\times\ldots\times V_m)$.

Then,

$$g(v_1,\dots, v_m)=\sum\limits_{i=1}^m\sum\limits_{j=1}^\infty a_{ij}w_{ij}\tag{2}$$

We now show the existence of an $(f_1,...,f_m)\in\mathcal{L}(V_1,W)\times\ldots\mathcal{L}(V_m,W)$ such that $\Gamma(f_1,\ldots,f_m)=g$.

Define $f_i\in \mathcal{L}(V_i,W)$ by

$$f_i(\alpha_{ij})=w_{ij}$$

for $i=1,\ldots,m$ and $j=1,\ldots,\infty$.

Then,

$$\Gamma(f_1,\ldots,f_m)=\sum\limits_{i=1}^m f_i$$

and

$$\Gamma(f_1,\ldots,f_m)(v_1,\ldots,v_m)=\Gamma(f_1,\ldots,f_m)\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j},\ldots \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )$$

$$=f_1\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j} \right )+\ldots+f_m\left ( \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )$$

$$=\sum\limits_{i=1}^m\sum\limits_{j=1}^\infty a_{ij}w_{ij}\tag{3}$$

Since (2) and (3) are equal we have $\Gamma(f_1,\ldots,f_m)=g$ and thus $\Gamma$ is surjective.

Since $\Gamma$ is a bijection between $\mathcal{L}(V_1\times\ldots\times V_m,W)$ and $\mathcal{L}(V_1,W)\times\ldots\times \mathcal{L}(V_m,W)$ then these vector spaces are isomorphic.

Answer 1

After the discussion in the comments it is clear that the proposed proof has some incorrect steps. It seems the proof can be saved, however, by correcting these steps.

Here is an attempt to correct them.

Let $\alpha_{i,1},\alpha_{i,2},\ldots,\alpha_{i,\infty}$ be a basis for $\mathcal{L}(V_i,W)$.

Define the linear map $\Gamma: \mathcal{L}(V_1,W)\times\ldots\times\mathcal{L}(V_m,W)\to \mathcal{L}(V_1\times\ldots\times V_m,W)$ by

$$\Gamma(f_1,\ldots,f_m)=\sum\limits_{i=1}^m \phi_{f_i}$$

where $\phi_{f_i}:V_1\times\ldots\times V_m\to W, (v_1,\ldots,v_m)\mapsto f_i(v_i)$.

Note that the above was the corrected portion. The sum $\sum\limits_{i=1}^m\phi_{f_i}$ is a sum of linear maps in $\mathcal{L}(V_1\times\ldots\times V_m,W)$.

In the original proposed proof, we had the sum $\sum\limits_{i=1}^m f_i$ which was summing vectors (ie, functions) from different vector spaces, namely, $\mathcal{L}(V_i,W)$ for $i=1,\ldots,m$.

Next, note that a basis for $\mathcal{L}(V_1\times\ldots\times V_m,W)$ is

$$(\alpha_{1j},0,\ldots,0)$$ $$\ldots$$ $$(0,\ldots,0,\alpha_{mj})$$

for $j=1,\ldots,\infty$.

Now suppose $\Gamma(f_1,\ldots,f_m)=\Gamma(h_1,\ldots,h_m)$.

Then, $\sum\limits_{i=1}^m\phi_{f_i}=\sum\limits_{i=1}^m\phi_{h_i}$.

Since

$$\phi_{f_i}=\sum\limits_{j=1}^\infty a_{1j}(\alpha_{1j},0,\ldots,0)+\dots+\sum\limits_{j=1}^\infty a_{mj}(0,\ldots,0,\alpha_{mj})$$

$$\phi_{h_i}=\sum\limits_{j=1}^\infty b_{1j}(\alpha_{1j},0,\ldots,0)+\dots+\sum\limits_{j=1}^\infty b_{mj}(0,\ldots,0,\alpha_{mj})$$

then

$$\sum\limits_{j=1}^\infty (a_{1j}-b_{1j})(\alpha_{1j},0,\ldots,0)+\dots+\sum\limits_{j=1}^\infty (a_{mj}-b_{mj})(0,\ldots,0,\alpha_{mj})=0$$

and because of linear independence we have that for each $i=1,\ldots, m$ we have $a_{ij}=b_{ij}$ for all $j=1,\ldots,\infty$.

Thus, $\phi_{f_i}=\phi_{h_i}$ for $i=1,\ldots,m$.

Thus, $f_i=h_i$ for $i=1,\ldots,m$.

Therefore, $\Gamma$ is injective.

Next we prove that $\Gamma$ is surjective. Again we need to correct the same error that happened in the original injection portion of the proof.

Let $g\in\mathcal{L}(V_1\times\ldots\times V_m,W)$.

Let $(v_1,\ldots,v_m)\in V_1\times\ldots\times V_m$ and

$$g(\alpha_{1j},0,\dots,0)=w_{1j}$$ $$\ldots$$ $$g(0,0,\dots,0,\alpha_{mj})=w_{mj}$$

for $j=1,\dots,\infty$.

Let $(v_1,\ldots,v_m)=\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j},\ldots \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )\in(V_1\times\ldots\times V_m)$.

Then,

$$g(v_1,\dots, v_m)=\sum\limits_{i=1}^m\sum\limits_{j=1}^\infty a_{ij}w_{ij}\tag{2}$$

We now show the existence of an $(f_1,...,f_m)\in\mathcal{L}(V_1,W)\times\ldots\mathcal{L}(V_m,W)$ such that $\Gamma(f_1,\ldots,f_m)=g$.

Define $f_i\in \mathcal{L}(V_i,W)$ by

$$f_i(\alpha_{ij})=w_{ij}$$

for $i=1,\ldots,m$ and $j=1,\ldots,\infty$.

Then (and this is the part that is corrected)

$$\Gamma(f_1,\ldots,f_m)=\sum\limits_{i=1}^m \phi_{f_i}$$

and

$$\Gamma(f_1,\ldots,f_m)(v_1,\ldots,v_m)=\Gamma(f_1,\ldots,f_m)\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j},\ldots \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )$$

$$=\phi_{f_1}\left ( \sum\limits_{j=1}^\infty a_{1j}\alpha_{1j} \right )+\ldots+\phi_{f_m}\left ( \sum\limits_{j=1}^\infty a_{mj}\alpha_{mj} \right )$$

$$=\sum\limits_{i=1}^m\sum\limits_{j=1}^\infty a_{ij}w_{ij}\tag{3}$$

Since (2) and (3) are equal we have $\Gamma(f_1,\ldots,f_m)=g$ and thus $\Gamma$ is surjective.

Since $\Gamma$ is a bijection between $\mathcal{L}(V_1\times\ldots\times V_m,W)$ and $\mathcal{L}(V_1,W)\times\ldots\times \mathcal{L}(V_m,W)$ then these vector spaces are isomorphic.

Answer 2

It might simplify notation, etc to start with the case $n=2$ and proceed by induction on $n$.

In fact the direct sum of two vector spaces $U\oplus V$ which you are writing as $U\times V$ has the following universal property: Given two linear mappings $f\colon U\to W$ and $g\colon V\to W$, there is a unique linear mapping $h\colon U\oplus V\to W$ such that $h\circ i=f$ and $h\circ j=g$ where $i\colon U\to U\oplus V$ and $j\colon V\to U\oplus V$ are the two inclusion maps. It is common to set $h=f\oplus g$.

This makes the result for $n=2$ obvious and the general case follows.

Actually it even works for an infinite collection of $V_i$ where $\bigoplus_iV_i$ is the collection of sequences with only finitely many non-zero terms. It does not depend on finite dimensionality of the $V_i$ or $W$.

This sort of result works in any abelian category (such as modules over a ring) because the finite direct sum is the coproduct (and also the product); if arbitrary sized coproducts exist then the general case works.