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If $f: A\to B$ is bijective, then it has an inverse $g: B \to A$ defined as
\begin{equation} g = \left \{ \big( f(a), a \big) : a \in A \right \} \end{equation} If $g: B \to A$ is any function for which,
\begin{equation} \begin{aligned} (f \circ g)(x) & = x, & \forall x \in B, \\ (g \circ f)(x) & = x, & \forall x \in A. \end{aligned} \end{equation} then $g \equiv f^{-1}$ (inverse)

I basically need to show that if any two functions are composed with $x$ ( I don't if that's the proper way to say it) and you get back $x$, then those two functions are inverses of one another.
In other words $g \equiv f^{-1}$ (inverse)

Any help with this would be really appreciated. I've been trying to get this down for a couple of days now.

Rodrigo Zepeda
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Marco Neves
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1 Answers1

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Well, if you have a set-theoretic definition of the inverse, it makes sense to expand the other two statements that you have in terms of set theory as well. In particular, we can expand the statement $$(f\circ g)(x)=x$$ to say that for any pair $(x,y)\in f$ we have that $(y,x)\in g$. This comes essentially directly from the definition of a composition of relations. The statement $$(g\circ f)(x)=x$$ is essentially the converse to this - saying that $(y,x)\in g$ implies $(x,y)\in f$.

Putting these together gives the following: $(y,x)\in g$ if and only if $(x,y)\in f$. Given that the latter condition is equivalent to saying $y=f(x)$ we get that $(y,x)\in g$ if and only if $y=f(x)$ - which immediately gives that $g=\{(f(x),x):x\in A\}$.

Milo Brandt
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  • That sort of makes sense. Except, I don't know if you can just assume that the converse of a statement implies what you said because you don't know if g is the inverse of f yet. – Marco Neves Nov 25 '15 at 22:29
  • What statement are you doubting exactly? Given that we have both $(x,y)\in f\Rightarrow (y,x)\in g$ and $(x,y)\in g\Rightarrow (y,x)\in f$ we have that those two conditions are equivalent and the rest is just rewriting this; nothing's been assumed. – Milo Brandt Nov 25 '15 at 22:34
  • Oh never mind, silly me. I get it now. Much thanks! – Marco Neves Nov 25 '15 at 22:39