Given a function a to b, its inverse relation will be a function iff the function is bijective

Question

So when I did this proof I didn't know I was supposed to physical prove all the parts out (assuming they are both functions and then using that to prove the inverse is an injection). My problem is I have never seen a relation proof using a function and an inverse function (only have done compositions so far). Can somebody explain how I format one part of this proof so I can get an understanding of how I use the definitions with an inverse function?

Answer 1

Recall that a relation between sets $X$ and $Y$ is simply a subset $R\subseteq X\times Y$. A function $f:X\to Y$ is a special relation $R_f\subseteq X\times Y$ such that for every $x\in X$, there is a unique $y\in Y$ for which $(x,y)\in R_f$ (this is precisely what we mean by $f(x)=y$). For a relation $R\subseteq X\times Y$, the inverse relation $\overline{R}\subseteq Y\times X$ is defined by $(y,x)\in\overline{R}$ iff $(x,y)\in R$.

So all that is being asked is to show that for a function $f:X\to Y$ (identified with its relation $R_f\subseteq X\times Y$) is a bijection iff $\overline{R_f}\subseteq Y\times X$ is a function (or if you prefer, is the relation associated with some function). Now just go forth and verify definitions!