Let (G1,∗) and (G2,◦) be groups. A mapping f : G1 → G2 is said to be a group homomorphism if for all g,h∈G1 f(g∗h)=f(g)◦f(h)
It's true that a function between groups preserving the group operation automatically preserves neutral elements. That's the reason why the condition is sometimes omitted. In this case, I want a group homomorphism G → GL(hom(V, W)). And if I already knew that the action of g∈G on hom(V, W) is invertible (so that it is an element of GL(hom(V, W))), then I would fine be. But how do you show that?
In image is my answer to original question, which I did, but I don't have the answer how I show that action is invertible.
When you say that you have a map from $G$ into $GL(\hom(V,W))$, you are already claiming that the image of each $g\in G$ is invertible.