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If f:A->B and g: B->A. Is this enough to say f and g are onto?
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Well it could be that \(\forall x\quad f(x) = b\in B\) and \(\forall x\quad g(x) =a\in A\). they'd both be constants and not onto.
|dw:1397792011372:dw|
wait.. hold on
oh I see it now,
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