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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? –  Adrián Barquero Dec 5 '10 at 19:36
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up vote 3 down vote accepted

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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To Prove Injectivity I would say that if –  Bradley Barrows Dec 5 '10 at 19:39
    
To Prove Injectivity could I say that since fof is the identity function then f(f(x)) = f(x) = x and f(f(y)) = f(y) = y. then x =y so its injective. To prove Surjectivity could I say that Given b∈A, there exists a∈A such that f(a)=b since, f(f(a))=id(a)=a so b=a? Also I was wondering If I could just say that since fof is an identity function it must be a composition of bijections therefore f is a bijection? –  Bradley Barrows Dec 5 '10 at 19:47
    
@Bradley: The injectivity part is right. For surjectivity, "proving" $b=a$ would mean that $f$ is identity which it need not be. Since, $f(f(a))=a$ what element maps to $a$ under $f$? –  Timothy Wagner Dec 5 '10 at 19:50
    
It seems like both a and f(a) map to a under f? My goal is to show b = f(a) correct? –  Bradley Barrows Dec 5 '10 at 20:07
    
So since f(f(a)) = a and f(b) = a then b = f(a)? –  Bradley Barrows Dec 5 '10 at 20:10
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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.

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