$\mathcal L(V_1 \times \cdots \times V_m,W) \approx \mathcal L(V_1,W) \times \cdots \times \mathcal L(V_m,W)$

Question

Suppose $V_1, \dots, V_m, W$ are vector spaces$^\dagger$. Prove that $\mathcal L(V_1 \times \cdots \times V_m,W)$ and $\mathcal L(V_1,W) \times \cdots \times \mathcal L(V_m,W)$ are isomorphic vector spaces.

I'm having some trouble with this one. If the vector spaces were finite-dimensional I could try proving that each has the same dimension, but the question doesn't specify that they're finite-dimensional spaces.

It seems like maybe I need to construct the isomorphism. So I'd need some linear mapping $$\big((v_1,\dots, v_m)\mapsto w\big)\mapsto (v_1 \mapsto w_1, \dots, v_m\mapsto w_m)$$ but TBH I'm getting a little confused from my own notation here.

Could someone give a hint as to how to get started on this exercise?

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^{$\dagger$: The running assumption is that we're talking about $\Bbb R$- or $\Bbb C$-vector spaces, if that's relevant.}

Answer 1

Hints:

• Do it for two factors first.

• If $f:V_1\times V_2\to W$ is a linear map, then you can consider the map $f_1:v\in V_1\mapsto f(v,0)\in W$ and, similarly, a map $f_2:V_2\to W$ constructed from $f$. In this way you can produce a function $$\Phi:f\in L(V_1\times V_2,W)\mapsto (f_1,f_2)\in L(V_1,W)\times L(V_2,W).$$ Show that it does what you want.

Answer 2

The natural linear surjections $$p_i:V_1\times\dots\times V_m\to V_i,\;(v_1,\dots,v_m)\mapsto v_i$$ and injections $$q_i:V_i\to V_1\times\dots\times V_m,\;u\mapsto(0,\dots,0,u,0,\dots,0)$$ (where $u$ is the $i$-th component in $(0,\dots,0,u,0,\dots,0)$) satisfy $$\sum_{i=1}^mq_i\circ p_i(v_1,\dots,v_m)=(v_1,\dots,v_m),\quad p_i\circ q_j(u)=\begin{cases}u&\text{if }i=j,\\0&\text{if }i\ne j. \end{cases}$$

Therefore, the two maps $$F:\mathcal L(V_1 \times \dots \times V_m,W)\to\mathcal L(V_1,W) \times \dots \times \mathcal L(V_m,W),\;f\mapsto(f\circ q_1,\dots,f\circ q_m)$$ $$G:\mathcal L(V_1,W) \times \dots \times \mathcal L(V_m,W)\to \mathcal L(V_1 \times \dots \times V_m,W),\;(f_1,\dots,f_m)\mapsto\sum_{i=1}^mf_i\circ p_i$$ are mutually inverse isomorphisms, since they are linear by construction, and satisfy $$(G\circ F)(f)=f,\quad(F\circ G)(f_1,\dots,f_m)=(f_1,\dots,f_m).$$