# If $f \circ f$ is bijective for $f: A \to A$, then is $f$ bijective?

I am trying to prove the following statement:

Let $$f: A \to A$$. If $$f \circ f$$ is a bijection, then $$f$$ is bijective.

My proof looked like this:

We know that $$|A| = |A|$$. Since this is the case, there exists a bijection $$f: A \to A$$ which has us conclude that $$f$$ is bijective.

Is this sufficient to prove the statement? Or must I separately prove surjectivity and injectivity for $$f$$ using $$f \circ f$$?

• Not exactly a duplicate, but almost: math.stackexchange.com/questions/13135/… – Asaf Karagila Mar 27 '20 at 8:24
• I don't think it's a duplicate because $f \circ f$ can be any bijection. Thanks for showing me the similar question though. – Kookie Mar 27 '20 at 8:26
• If I thought it was a duplicate, I would have closed the question. It's almost a duplicate, since the argument is quite similar. – Asaf Karagila Mar 27 '20 at 8:27
• Oh, right. That would make sense then. :P – Kookie Mar 27 '20 at 8:27
• There exists a bijection $A\to A$, but not necessarly $f$!! – Jean-Claude Colette Mar 27 '20 at 8:28

## 3 Answers

You are asked to prove a property of a particular function $$f$$. Saying that there exists a bijection From $$A$$ onto $$A$$ proves nothing.

Suppose $$f\circ f$$ is a bijection. The $$f(x)=f(y)$$ implies $$f(f(x))=f(f(y))$$ which implies $$x=y$$. So $$f$$ is injective. I will let you show that $$f$$ is surjective also.

• Can I prove surjectivity by saying that since $f \circ f$ is surjective, $\forall x \in A, \exists y \in A$ such that $f \circ f(x) = y$, then let $k = f(x) \in A$ and say $\forall k \in A, \exists y \in A$ such that $f(k) = y$ and conclude that $f$ is surjective? – Kookie Mar 27 '20 at 8:37
• @Kookie Yoo wrote $\forall x \in A \exists y \in A$ instead of $\forall y \in A \exists x \in A$ – Kavi Rama Murthy Mar 27 '20 at 8:40
• It's not a typo. Is it really the other way around? – Kookie Mar 27 '20 at 8:42
• If you say for each $x$ there exists $y$ such that $f(x)=y$ you are only saying that $f$ is a well defined function. To show that it is surjective you have to show that for each $y$ there exists $x$ such that $f(x)=y$. – Kavi Rama Murthy Mar 27 '20 at 8:47

Hint: if $$fo g$$ is bijection $$\implies g$$ is one-one and $$f$$ is onto.

Your proof comes down to using the same letter $$f$$ for two different things. You are, first of all, given an $$f: A \longrightarrow A$$. You furthermore write that there exists a bijection $$A \longrightarrow A$$. You shouldn't call this map $$f$$, as there's no reason to believe that it's the same as the map you're originally. Indeed, your argument would imply that all maps $$A \longrightarrow A$$ are bijections, which is nonsense if $$|A| \geq 2$$.

Back to the proof itself, we do indeed have to show that $$f$$ is injective and surjective. We certainly have $$im(f \circ f) = A$$. Furthermore, $$im(f \circ f) = f[f[A]]$$. Hence, $$im(f) \supseteq f[f[A]] = A$$, so $$f$$ is onto. For injectivity, suppose $$f(x) = f(y)$$. By surjectivity, let $$f(a) = x$$, $$f(b) = y$$. Then $$f(f(a)) = f(x) = f(y) = f(f(b))$$, so $$a = b$$. Thus, $$x = f(a) = f(b) = y$$.