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Ok so this is a long one but I want to make sure I understand well.

Assume we know that $f:X \rightarrow Y$ is invertable iff it is a bijection and we know that the inverse is also a bijection as well as the fact that inverse functions are unique.

Exercise: Show given $f:A \rightarrow B $ and $g:B\rightarrow C $, bijections and given that both $f^{-1} \ \text{and} \ g^{-1}$ then $$(g\circ f)^{-1} = f^{-1} \circ g^{-1}$$

Claim: Let $f:A \rightarrow B $ be bijective, then $ (f^{-1})^{-1} = f$

Let $g = f^{-1} $. By a theorem seen in class, g is bijective and also has an inverse. Moreover, $g:B\rightarrow A $

let $h=g^{-1}$ then by definition we get that $h:A \rightarrow B$ and h bijective.

since $h$ is the inverse of $g$ then we get the following results

$$g\circ h = ID_{A} \ \text{and} \ h\circ g = ID_{B}$$

since $g$ is the inverse of $f$ it also follows that

$$g\circ f = ID_{A} \ \text{and} \ f\circ g = ID_{B}$$

then $h=f$ since inverses are unique and thus $(f^{-1})^{-1} = g^{-1} = h = f$

Corollary: if $f$ is a bijection and $g$ is its inverse then $g^{-1} = f$

This follows directly from the proof above.

Proof of Exercise:

By the above corollary we only have to show that $(g\circ f)$ is the inverse of $f^{-1} \circ g^{-1}$

$(g\circ f) \circ (f^{-1} \circ g^{-1})$

$g\circ f \circ f^{-1} \circ g^{-1}$ by associativity of composition

$g\circ ID_{B} \circ g^{-1}$

$g\circ g^{-1}$

$ID_{C}$

and

$(f^{-1} \circ g^{-1}) \circ (g\circ f)$

$f^{-1} \circ g^{-1} \circ g\circ f$ by associativity of composition

$f^{-1} \circ ID_{B} \circ f$

$f^{-1} \circ f$

$ID_{A}$

We have shown that $(g\circ f)$ is the inverse and so $(g\circ f)^{-1} = f^{-1} \circ g^{-1}$

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    $\begingroup$ yep it's legit, is there any particular reason why you thought it would not be? $\endgroup$
    – mdave16
    Commented Sep 15, 2017 at 23:08
  • $\begingroup$ This same proof works in any algebraic structure, as long as the concept of inverses makes sense, inverses exist for $f$ and $g$, and the algebraic operation applied to $f$ and $g$ makes sense. Function composition is one example, and so is for instance matrix multiplication $\endgroup$
    – Arthur
    Commented Sep 15, 2017 at 23:13
  • $\begingroup$ Yeah, idk I'm never too sure with elementary proofs using definitions and whatnot. I feel weird proving things that are obvious to me so I never know if it's wrong or not. $\endgroup$ Commented Sep 16, 2017 at 21:13

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I don't see anything wrong with it. It depends somewhat on what your definition of $f^{-1}$ is, though. Given a bijection $f:A\to B,$ some texts would say that $f^{-1}$ is a function $g:B\to A$ satisfying $f\circ g=id_B$ and $g\circ f=id_A.$ One can readily prove that such a function $g$ is unique, once one has proved that it must exist if $f$ is a bijection. If that is your definition of $f^{-1},$ though, then there is no need for your Claim (nor its Corollary, which says exactly the same thing, in different words). We only need to show that $f^{-1}\circ g^{-1}$ is the inverse of $g\circ f,$ which is exactly what you showed in your final proof.

I must also say that I'm curious about "a theorem seen in class." I'd expect that such a theorem would already have shown that $f=\left(f^{-1}\right)^{-1},$ but it seems that it must not have done so, or you'd not have made your Claim.

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  • $\begingroup$ The theorem was $f$ has an inverse iff $f$ is bijective $\endgroup$ Commented Sep 16, 2017 at 23:09

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