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I've already proven that if we assume f is bijective and g is bijective, then $(g \circ f)$ is bijective. I've also proven that$(g \circ f)^{-1}$ exists. I'm stuck on this part, however. Any suggestions?

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Do you know the rule for groups? The proof here is the same. – Neal Dec 4 '12 at 3:46

Just follow the definitions.

Suppose that $\langle x,y\rangle\in(g\circ f)^{-1}$; clearly $\langle y,x\rangle\in g\circ f$, so by the definition of composition there must be some $z$ such that $\langle y,z\rangle\in f$ and $\langle z,x\rangle\in g$. But then $\langle z,y\rangle\in f^{-1}$ and $\langle x,z\rangle\in g^{-1}$, so again by the definition of composition we have $\langle x,y\rangle\in f^{-1}\circ g^{-1}$, and hence $(g\circ f)^{-1}\subseteq f^{-1}\circ g^{-1}$. The opposite inclusion is proved similarly: all of the steps in the argument are reversible.

Alternatively, if you’re not working at this low a set-theoretic level, just verify that

$$\left(f^{-1}\circ g^{-1}\right)\circ(g\circ f)$$ is the identity function on the domain of $f$, and

$$(g\circ f)\circ\left(f^{-1}\circ g^{-1}\right)$$ is the identity function on the range of $g$. These are trivial calculations if you know that composition of functions is associative.

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The symbol $(g \circ f)^{-1}$ reads: the function that, when composed with $g \circ f$, gives the identity. This is a description that uniquely characterizes $(g \circ f)^{-1}$, so you only need to check that $f^{-1} \circ g^{-1}$ has this property.

Also, don't forget to check compositions on both sides!

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Thank you. I had forgotten that $(g \circ f)^{-1}$ is unique. Now that I know that the solution is simple. – Bill Dec 4 '12 at 3:59

Recall that $(g\circ f)(a)=g(f(a))$. Three very useful statements which I will suppose you have proved before, at one point or another, are:

  1. The composition of injective functions is injective.
  2. The composition of surjective functions is surjective.
  3. If $f$ is a bijection then $f^{-1}$ is a bijection and $(f^{-1}\circ f)(a)=a$.

(If you haven't seen any of these before, please leave a comment and I'll add some hints about how to prove them, but it's really an exercise of unwinding the definitions)

So we assumed that $f\colon A\to B$ and $g\colon B\to C$ are bijections, by $(1), (2)$ we have that $g\circ f$ is also a bijection, and by $(3)$ we have that $(g\circ f)^{-1}$ is a bijection as well, and $f^{-1}\circ g^{-1}$ is also a bijection. Note that $\operatorname{dom}\Big((g\circ f)^{-1}\Big)=\operatorname{dom}\Big(f^{-1}\circ g^{-1}\Big)=C$.

To show that two functions with the same domain are equal we only need to show that they do the map every element the same. Namely we need only to verify that for every $c\in C$ the following equality holds: $$(g\circ f)^{-1}(c)=(f^{-1}\circ g^{-1})(c)$$

Let $a$ denote $(g\circ f)^{-1}(c)$, then we know that $g(f(a))=c$. We also know that $(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(c))$. But $g$ is a bijection therefore $g^{-1}(c)=f(a)$ from the two things we know here. But $f$ is a bijection and by $(3)$ we have that $f^{-1}(f(a))=a$, so finally we have: $$(f^{-1}\circ g^{-1})(c)=f^{-1}(g^{-1}(c))=f^{-1}(f(a))=a=(g\circ f)^{-1}(c)$$

And that what we wanted to show.

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