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 $f=g$?
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 $f_i\neq 0$.