We have a surjective function $f:A\rightarrow A$ and we know that $f\circ f=f$.
How do I prove that $f=id_A$?
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Let $x \in A$. Then since $f$ is surjective, there exists a $y \in A$ such that $$f(y)=x$$
This implies $$f(f(y))= f(y)$$
$$f(x)=x$$
Since our choice of $x$ was arbitrary, we're done.
Zubin Mukerjee
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I don't underrstand your final step – bolzanoman Jan 15 '15 at 20:48
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It uses what he wrote down was a result of $f$ being both a self-map and surjective. – Arturo don Juan Jan 15 '15 at 20:52
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@ZubinMukerjee could you explain your funal step a bit more? – bolzanoman Jan 15 '15 at 20:52
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@arturodonjuan but he cant use the fact that f i a selfmap, since thats what we need to prove – bolzanoman Jan 15 '15 at 20:54
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1@bolzanoman There are so few steps that it is hard to see what is meant with final step.- Given $x\in A$, we want to show that $f(x)=x$. By surjectivity of $f$, there exists $y$ with $x=f(y)$. Then by this and the property of $f$, we have $f(x)=f(f(y))=(f\circ f)(y)=f(y)=x$ as desired. – Hagen von Eitzen Jan 15 '15 at 20:55
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1@bolzanoman he substituted in the first line, $f(y)=x$. – jh4 Jan 15 '15 at 20:55
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1@everybody I'm stupid! Only now understand Thanks a bunch! – bolzanoman Jan 15 '15 at 20:56