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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?

George
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  • Hint: If the function $f:A\rightarrow B$ is bijective, for any $b\in B$ there exists (from surjectiveness) a unique (from injectiveness) $a\in A$ with $f(a)=b$. What does this mean about the domain of $f^{-1}$? – NL1992 Dec 03 '18 at 04:44

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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!

munchhausen
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  • ohh ok now I get it, but to make sure can you tell me if what I wrote for the injection part sound right?---------fix $(x,y_1),(x,y_2)\in f$ for some $x\in A$ and fix $y_1,y_2\in B$. Then by the assumption that $f^{-1}$ is a function from $B$ to $A$. $(x_1,y),(x_2,y)\in f^{-1}$ for some $y\in A$. Since $(x_1,y),(x_2,y)\in f^{-1}$ this impiles $(x_1=x_2)$. Thus by definition of injective relations, function $f$ is an injection. – George Dec 03 '18 at 05:04
  • here what that definition states----- A relation $R\subseteq A X B$ is injective if $(x_1,y) \in R$ and $(x_2,y) \in R$ implies $x_1=x_2$ – George Dec 03 '18 at 05:11
  • So, you're getting things flipped around. $(x,y)\in f$ is the statement that $f(x)=y$. So to prove that it $f^{-1}$ is a function, then $f$ is injective, you want to start by assuming that $(x_1,y),(x_2,y)\in f$ (so that we have $f(x_1)=f(x_2)=y$). Then you know that $(y,x_1),(y,x_2)\in f^{-1}$, but since $f^{-1}$ is a function, by assumption, this tells you that $x_1=x_2$. – munchhausen Dec 03 '18 at 06:12