# Thread: Proof of being onto given a composition is onto?

1. ## Proof of being onto given a composition is onto?

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:

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)).
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)). Since the composition of g with f maps to C, then f must be onto.
But I'm not too confident in that.

2. Originally Posted by mjlaz
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:

But I'm not too confident in that.
Suppose that $\displaystyle f$ was not onto, then since $\displaystyle f(A)\subseteq B$ we see that $\displaystyle g(f(A))\subseteq g(B)\subset C$