Suppose that g: A -> B and f: B-> C and that f o g is onto. Show that f is onto.
Here's what I have written on my pad:
The next step isn't quite coming to me so easily. I want to say this:Proof: Suppose g: A -> B and f: B-> C and that f o g is onto. Observe that f o g: A -> C. Since we have that f o g is onto, then C = (f o g)(A) = f(g(A)).
But I'm not too confident in that.Proof: Suppose g: A -> B and f: B-> C and that f o g is onto. Observe that f o g: A -> C. Since we have that f o g is onto, then C = (f o g)(A) = f(g(A)). Since the composition of g with f maps to C, then f must be onto.