Suppose $V_1, \dots, V_m$ are vector spaces. Prove that $\mathcal{L}(V_1 \times \dots \times V_m, W)$ is isomorphic to $\mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W).$ (Note that $V_{i}$'s can be infinite-dimensional.)
I am having trouble showing that $\varphi$ defined below is surjective. For every $f \in \mathcal{L}(V_1 \times \dots \times V_m, W),$ I defined $f_{i}: V_{i} \to W$ by $$f_{i} (v_{i}) = f (0, \dots, v_{i}, \dots, 0).$$ Then, I defined $\varphi: \mathcal{L}(V_1 \times \dots \times V_m, W) \to \mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W)$ by $$\varphi (f) = (f_{1}, \dots, f_{m}).$$
Now, how would I show that $\varphi$ is surjective?
I know I have to show that for any $(g_{1}, \cdots, g_{m}) \in \mathcal{L}(V_1, W) \times \dots \times \mathcal{L}(V_m, W)$, there is a corresponding $g \in \mathcal{L}(V_1 \times \dots \times V_m, W)$ so that $\varphi (g) = (g_{1}, \dots, g_{m}).$
Can I simply define $g \in \mathcal{L}(V_1 \times \dots \times V_m, W)$ by
$$g (0, \dots, v_{i}, \dots, 0) = g_{i} (v_{i})? $$
I am not sure where to start.
For $(g_1,\dots,g_m) \in \mathcal{L}(V_1,W) \times \cdots \times \mathcal{L}(V_m,W)$ define $g : V_1 \times \cdots \times V_m \to W$ by $$g(v_1,\dots,v_m) = g_1(v_1) + \cdots + g_m(v_m).$$