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I've constructed a linear map $\Psi$ that I think is an isomorphism. Let $V, W_1, \dots, W_m$ be vector spaces. Then first we define a projection function $$\pi_i: W_1 \times \cdots \times W_m \to W_i \\ \pi_i(w_1, \dots, w_m) = w_i$$ for $i = 1, \dots, m$. Then $\Psi$ is given by $$\Psi: \mathcal L(V, W_1\times \cdots \times W_m) \to \mathcal L(V, W_1)\times \dots \times \mathcal L(V, W_m) \\ \Psi(f) = (f_1, \dots, f_m)$$ where $f_i := \pi_i \circ f$.

I've already proved that it's linear. So to prove that it's an isomorphism I need to show that it's invertible or equivalently that it's bijective. To prove that I first prove that it's injective. That's where I'm getting stuck. Here's what I've got:

Suppose $\Psi(f) = \Psi(g)$. Then $$(\pi_1 \circ f, \dots, \pi_m \circ f) = (\pi_1 \circ g, \dots, \pi_m \circ g),$$ i.e. $\pi_i \circ f = \pi_i \circ g$. From here I can't think what argument I can use to strip the $\pi_i$ off. It's clearly not invertible itself because it's a nontrivial projection. So how do I go from here to show that $f=g$?

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    In essence, it isn't different from the fact that $(v_1,\cdots,v_n)=(w_1,\cdots,w_n)$ if and only if $v_i=w_i$ for all $i$ – 2017-02-15

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If $\psi$ is linear, then $\psi$ is injective $\iff$ $\ker \psi=0$. So, if $f\in \ker \psi$, that means $(f_1, ..., f_m)=0$, which is the same as $f_i(v)=0 \ \forall v\in V$ and for all $i\in\{1, \dots, n\}$. Now, we would like to prove that this implies $f=0$. If $f\neq 0$, there exists $v\in V$ so that $f(v)=(w_1, \dots, w_n)\neq 0$, which means that at least one of the $w_i$ has to be different from $0$. That implies $f_i(v)=(\pi_i \circ f)(v)=w_i\neq 0$, so $f_i\neq 0$.

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    The problem with that is that I think I've only proven that first statement for finite dimensional spaces. But let me look back through my notes and check. – 2017-02-15
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    I don't remember you need to assume finite dimensions to prove that, but it's a good idea to check it up. – 2017-02-15
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    Nope, I don't need finite dimensions. I proved it for all linear maps, apparently. Thanks! – 2017-02-15
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    Actually, what you were trying to prove, $\psi(f)=\psi(g)\Rightarrow f=g$, is the same as trying to prove that $\psi(h)=0 \Rightarrow h=0$, because if you have proven the latter, then letting $h=f-g$ you get the former. – 2017-02-15