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 $f \circ f$?
Best Answer
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.