I am trying to study functions in math and learning some basic proofs. In numerous places I have seen this: $$(f \circ\ g) ^{-1}(u) = g^{-1}(f^{-1}(u))$$ I know this is true as well, having used it in numerous places in middle and high school. Is there any way of proving this definition though using logical steps? Thank you!
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Cf. this – J. W. Tanner Feb 20 '20 at 03:42
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1Think of $g$ as putting on socks, and $f$ as putting on shoes. Then $g^{-1}$ would be taking off socks, and $f^{-1}$ would be taking off shoes. $f\circ g$ would be putting on socks and then putting on shoes. $(f\circ g)^{-1}$ would be undoing the process of putting on socks and shoes. Hence $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$, because to undo the process of putting on socks and shoes, you must remove shoes first and then remove socks second. This is a theorem from topology. – user729424 Feb 20 '20 at 04:04
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That analogy, though funny, makes so much sense. Thanks a lot! In fact, im doing topology right now. Could you tell me what theorem exactly are you talking about? Did you mean what I asked is a theorem? – gazillionTickets Feb 20 '20 at 04:08
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I was joking a bit when I said that the fact you must take off your shoes before you take off your socks was a theorem of topology. I do think topology is what you would use if you had to prove this sort of thing. In any event, my wife pointed out that the "theorem" wouldn't work on me since my socks are not homotopy equivalent to normal socks because of the holes in them. – user729424 Feb 20 '20 at 04:18
4 Answers
Assuming that $g : A \to B$ and $f : B \to C$, then, by definition, $(f\circ g)^{-1} : C\to A$ is the unique function such that
- $(f\circ g)^{-1}\circ(f\circ g) = \textrm{id}_A$, and
- $(f\circ g)\circ(f\circ g)^{-1} = \textrm{id}_C$.
But observe that the function $g^{-1}\circ f^{-1} : C\to A$ has the same properties: $$\begin{align} (g^{-1}\circ f^{-1}) \circ (f\circ g) &= \big( g^{-1} \circ (f^{-1} \circ f) \big) \circ g \\ &= ( g^{-1} \circ \textrm{id}_B) \circ g \\ &= g^{-1} \circ g = \textrm{id}_A \end{align}$$ and $$\begin{align} (f\circ g) \circ (g^{-1}\circ f^{-1}) &= \big( f \circ (g \circ g^{-1}) \big) \circ f^{-1} \\ &= ( f \circ \textrm{id}_B) \circ f^{-1} \\ &= f \circ f^{-1} = \textrm{id}_C. \end{align}$$ Hence, $g^{-1}\circ f^{-1}$ and $(f\circ g)^{-1}$ are the same!

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@AbdulKalam It is the identity function on $A$: the function that takes each $a \in A$ into itself. – azif00 Sep 20 '20 at 01:27
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I didn't understand, can't you show A directly without using identity function? – Abdul Kalam Sep 20 '20 at 02:09
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Like, why to use identity function at all, why not just depict f^-1(f(X))= X instead of idX? – Abdul Kalam Sep 20 '20 at 02:43
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@AbdulKalam The statement "$f^{-1}(f(x)) = x$ for all $x \in B$" is equivalent to "$f^{-1} \circ f = \operatorname{id}_B$". This is what you mean? Using elements and quantifiers instead of equality of functions? – azif00 Sep 20 '20 at 03:30
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Assume functions $g:A\mapsto B, f:B\mapsto C$ are bijective (therefore invertible).
By definition : $(f\circ g)(x)=f(g(x))$
Applying inversion:
$$g^{-1}(f^{-1}((f\circ g)(x)))=x$$
Let : $x=(f\circ g)^{-1}(u)$ which means also $u=(f\circ g)(x)$
Therefore …$$g^{-1}(f^{-1}(u))=(f\circ g)^{-1}(u)$$

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To prove something is the inverse of a function, you need only check the compositions work as expected, i.e. if you believe $g(x)$ is the inverse of $f(x)$, then show $f(g(x))=x$ and $g(f(x))=x$. This will prove that $g(x)= f^{-1}(x)$.
Define $G(u)= (f \circ g)(u)$ and $H(u)= g^{-1}(f^{-1}(u))$. Now just check that $H(G(u))=u$ and $G(H(u))=u$. You will want to use the fact that $f,g$ have inverses, i.e. $f^{-1}, g^{-1}$ exist so that $f(f^{-1}(x))=x$, $f^{-1}(f(x))=x$ and $g(g^{-1}(x))=x$, $g^{-1}(g(x))=x$.

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For any $x$ in the domain, let $y=(f\circ g)^{-1}(x)$. Note that the following are equivalent:
$$y=(f\circ g)^{-1}(x)$$ $$(f\circ g)(y)=x$$ $$f(g(y))=x$$ $$g(y)=f^{-1}(x)$$ $$y=g^{-1}(f^{-1}(x))$$ $$y=(g^{-1}\circ f^{-1})(x)$$
It follows that $(f\circ g)^{-1}(x)=y=(g^{-1}\circ f^{-1})(x)$

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