The problem requires me to show that the following statement is true or false.
Let $A,B$ be sets, and $f:A\to B$, $g:B\to A$ be functions. Suppose $g\circ f\circ g$ is surjective, and $f\circ g\circ f$ is injective. Then $f\circ g$ is bijective.
So I get:
“$g\circ f\circ g$ is surjective and $f\circ g\circ f$ is injective“ $\implies$ “$g\circ f$ is bijective“ $\implies$“$g$ is surjective and $f$ is injective”
But since it does not imply that “$f\circ g$ is bijective”, I try to find some counter examples. But it is quite difficult.
Normally, if the question turns to be:
Let $A,B$ be sets, and $f:A\to B$, $g:B\to C$ be functions. Suppose $g$ is surjective, and $f$ is injective. Then $f\circ g$ is bijective.
I can easily disprove it since $A={0,1}, B=C={2}, f(2)=2,g(0)=2,g(1)=2$ is already a counter example.
But in this case, the set $B$ has to map back to $A$ by $g$. Then seems I can never find a counter example? Is it actually bijective? Or what else did I miss?
