Invertible compostion of 3 functions

Question

Let $f,g,h$ be functions from $A$ to $A$. Assume the composition $f \circ g \circ h$ is invertible. Is $g$ injective? Is $g$ on $A$? This seems to be true if $A$ is a finite set, but is it true if $A$ is not a finite set?

Answer 1

Let $A=\{(a_1, a_2, \ldots ) \, | \, a_i \in \mathbb{R}\}$ be the set of sequences. Consider \begin{align*} f(a_1, a_2, \ldots )&=(a_2, a_3, \ldots )\\ g(a_1, a_2, \ldots )&=(0,a_1, a_2, \ldots )\\ h(a_1, a_2, \ldots )&=(a_1, a_2, \ldots ) \end{align*} Then $f\circ g\circ h$ is the identity function, hence invertible. But $g$ (not onto) and $f$ (not one-one) are not.

Note: When $|A|< \infty$, then if a function $k:A \longrightarrow A$ has one of the two among injectivitiy and surjectivity, then it has the other as well. So, it works in the finite case.