Inverse of a Bijection

Let $X$ and $Y$ be sets, and write $\mathrm{id}_{X}:X\to X$ and $\mathrm{id}_{Y}:Y\to Y$ for the identity maps, given by $\mathrm{id}_{X}(x)=x$ and $\mathrm{id}_{Y}(y)=y$. For maps $f:X\to Y$ and $g:Y\to X$ let $g\circ f:X\to X$ be the map with $(g\circ f)(x)=g(f(x))$, and let $f\circ g:Y\to Y$ be the map with $(f\circ g)(y)=f(g(y))$. The notion bijection is that of the indicated definition.

Then the following hold.

1. (Existence and uniqueness of an inverse) If $f:X\to Y$ is a bijection, then there is exactly one map $g:Y\to X$ with

$$g\circ f=\mathrm{id}_{X}\qquad\text{and}\qquad f\circ g=\mathrm{id}_{Y}.$$

This map is written $f^{-1}$.

2. (The inverse is a bijection) If $f:X\to Y$ is a bijection, then $f^{-1}$ is a bijection from $Y$ to $X$, and $(f^{-1})^{-1}=f$.

3. (Two-sided inverses detect bijections) If $f:X\to Y$ and $g:Y\to X$ satisfy $g\circ f=\mathrm{id}_{X}$ and $f\circ g=\mathrm{id}_{Y}$, then $f$ and $g$ are bijections and $g=f^{-1}$.