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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.2012-03-05

1 Answers 1

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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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    $a \neq b$, right?2012-03-05
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    @The Chaz: absolutely! - thanks for that.2012-03-05
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    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 :)2012-03-05
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    I 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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    @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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    So 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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    Huzzah! Thank you for the help!!2012-03-05