# Prove: If $g\circ f$ is $1-1$ and $f$ is onto, then $g$ is $1-1.$

Let $f:A \to B, g:B \to C$. I don't really know how to prove this but I understand what it means.

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As a general hint with these types of problems see what happens if you assume the conclusion is false. –  Chris Janjigian Mar 5 '12 at 4:54

Assume not.

Then there exist $a,b$ with $a \neq b$ s.t. $g(a) = g(b)$. $f$ onto means there exist $c,d$ s.t. $f(c) = a, f(d) = b$. But then...

Does that help?

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$a \neq b$, right? –  The Chaz 2.0 Mar 5 '12 at 5:26
@The Chaz: absolutely! - thanks for that. –  mixedmath Mar 5 '12 at 5:30
It's like the comment (MSE link?) about omitting the dx in an integral: you and I know the condition is there without writing it, but it might confuse a new student of functions :) –  The Chaz 2.0 Mar 5 '12 at 5:35
I understand that you're assuming that g is not 1-1 first, right? But I don't know what follows after that... –  Andrea Douglas Mar 5 '12 at 5:59
Huzzah! Thank you for the help!! –  Andrea Douglas Mar 5 '12 at 6:21