Functions. Basics

Let's see that $f$ is an injective function. Suppose that $f(x) = f(y) \Rightarrow x = Id_{A} (x) = g \circ f(x) = g(f(x)) = g(f(y)) = g \circ f(y) = Id_{A}(y) = y \Rightarrow$ $f$ is an injective function.( Other way: if you suppose that $x \neq y$ then $f(x) \neq f(y)$ using reductio ad absurdum). Now, due to $Id_{A}$ is a surjective function $g$ is a surjective function. Thus, given $x \in A$ , then $x = Id_{A} (x) = g \circ f(x) = g(f(x)) \Rightarrow$ $\exists y \in B$ such that $g(y) = x$ .

There is no more possibilities... Consider $f : \{1,2\} \to \{1,2,3\}$ such that $f(x) = x \space \forall x \in \{1,2\}$ and $g:\{1,2,3\} \to \{1,2\}$ such that $g(1) = 1, g(2) = 2, g(3) = 1$ , then $g \circ f = Id_{A}$ and $f$ is neither surjective nor bijective and $g$ is neither injective nor bijective.