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

Question

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:

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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.

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

Answer 1

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.

Answer 2

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, $x = f(a) = f(b) = y$.

Answer 3

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