Let $f:A \to B, g:B \to C$. I don't really know how to prove this but I understand what it means.
Prove: If $g\circ f$ is $1-1$ and $f$ is onto, then $g$ is $1-1.$
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0As a general hint with these types of problems see what happens if you assume the conclusion is false. – 2012-03-05
1 Answers
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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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0$a \neq b$, right? – 2012-03-05
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0@The Chaz: absolutely! - thanks for that. – 2012-03-05
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0It'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 :) – 2012-03-05
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0I understand that you're assuming that g is not 1-1 first, right? But I don't know what follows after that... – 2012-03-05
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0@Andrea: You have to get the contradiction, because it's a proof by contradiction. So think to yourself: what is the contradiction we're working towards? There is only one step more to make. – 2012-03-05
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0So then g(f(c))=g(f(d)) which means that a=b, which is a contradiction. Is that correct? – 2012-03-05
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1Huzzah! Thank you for the help!! – 2012-03-05