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Let $f\colon A\to A$; prove that if $f\circ f = \mathrm{id}_A$ then $f$ is a bijection.

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    Did you try anything at all?2010-12-05

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Hints: Injectivity: Suppose $f(x)=f(y)$, what can you say about $f(f(x))$ and $f(f(y))$. Surjectivity: Given $a\in A$, there exists $b\in A$ such that $f(b)=a$ since, $f(f(a))=id(a)=a$ so b=?

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    @Bradley: Your last comment is right. Your second last comment is not necessarily true. $a$ may not map to $a$, but $f(a)$ does.2010-12-05
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This is a job for the Green and Brown Fact, namely Theorem 8 on page 14 of these notes. (The name comes from the colors of the chalk I used to draw a box around it that day in class.) If you have a composition of functions $g \circ f$, this tells you that if the composite function is injective, $f$ is injective, and if the composite function is surjective, $g$ is surjective.

The question asked by the OP fits so nicely into this framework that I feel embarrassed not to have discussed it either time I taught the course. Next time I will.

Ooh, another good exercise: what if $f \circ f \circ f = \operatorname{Id}_A$, and so forth.