Problem: Let $A$ and $B$ be two sets and $f:A\to B$ and $g:B\to A$ two functions such that $f\circ g\circ f$ is a bijection. Prove that $f$ and $g$ are also bijection.
Proof Attempt: We first show that $f$ is an injection. Let $f(x)=f(y)$ then $$g(f(x))=g(f(y))$$
$$f(g(f(x)))=g(g(f(y)))$$
$$(f\circ g\circ f)(x)=(f\circ g\circ f)(y)$$
$$x=y.$$
So it is an injection.
Now we will show $f$ is a surjection. Let $y\in B$ then there exists an $x\in A$ such that $(f\circ g\circ f)(x)=y.$ Or in other words for any $y\in B$ there exists an element $x'=g(f(x))\in A$ such that $f(x')=y.$ Hence $f$ is a surjection. Thus $f$ is a bijection.
Now we will show that $g$ is a bijection. We know that $f$ is a bijection and hence $f^{-1}$ is a bijection so we have that $f^{-1}\circ f\circ g\circ f\circ f^{-1}$ is a bijection. But this mapping is identical to $g$ and so we have that $g$ is a bijection.
Is this a valid proof?
Best Answer
Just to avoid this question to stay unanswered: it is a valid proof.