Condition for being invertible and one-to-one

Question

Consider $f(x)$ and $g(x)$ are two functions . When we can say composition of them (i.e. $f\circ g$ and $g\circ f$) is one-to-one and we can define inverse function ? If $f$ being one-to-one (in $f\circ g$ case) is sufficient or also $g$ have to be one-to-one ?

Answer 1

Consider the following cases (assume that the domain for both $f$ and $g$ contain at least two elements):

1) $f$ and $g$ are one-to-one then $f\circ g$ is one-to-one. Choose $x,y$ with $x\ne y$. Then,$g(x)\ne g(y)$ by the injectivity of $g$. Finally, by the injectivity $f(g(x))\ne f(g(y))$. Hence, $f\circ g$ is injective.

2) $f$ is one-to-one but $g$ is not then $f\circ g$ is not necessarily one-to-one. Indeed, choose $g(x)=0$ for all integers $x$ and $f(x)=x$ for $x\in\mathbb{N}$.

3) $f$ is not one-to-one and $g$ is one-to-one then $f\circ g$ is not one-to-one. Indeed, choose $f(x)=0$ and $g(x)=x$ for all $x\in\mathbb{N}$

For the first statement: in order to have an inverse $f\circ g$ must be a bijection. This implies that $f$ is surjective and $g$ injective (but nothing more wrt injectivity or surjectivity).

Hope this answers your doubt.