Discrete Math composition functions

Question

Give examples of sets $X$, $Y$, $Z$ and functions $f: X \to Y$, $g: Y \to Z$, so that the composition $g\circ f: X \to Z$ is a bijection, although neither $f$ or $g$ it is.

I have no idea to begin thinking on amounts.

Answer 1

Let $X=Z=\{1\}$ and $Y=\{1,2\}$. Now take $f$ injective but not surjective: $f(1)=1$. Moreover take $g$ surjective but not injective: $g(1)=g(2)=1$. Verify that $g\circ f:X\to Z$ is a bijecton.

Answer 2

WLOG we can take $Z = X$ and $g \circ f = Id$

If you can apply $f$ without "losing information", it's because $f$ is bijective on its image, i.e. injective. Thus take Y bigger than X. Then you just have to left invert $f$.

If you want to take $Z \neq X$ and $g \circ f \neq Id$, just left compose with a bijection from $X$ to $Z$.