Given A,B,C are sets and g: A -> B, f: B -> C.
I've been at this for a while now while also focusing on similar problems. I'm thinking a contradiction what be easiest suited.
Suppose f o g is surjective, g is not surjective and with the view to contradiction, suppose f is injective.
Since g is not surjective, we know there exists some b in B such that for all a in A, g(a) != b.
Moreover, since f is injective, for all b, b' in B such that b != b' => f(b) != f(b').
let c be in C, such that c = f(b), but there exists no g(a) = b, and given there exists a c that has no a such that (f o g)(a) = c, this opposes our assumption that f o g is surjective.
This is a contradiction, so f is not injective.