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I'm doing exercise on discrete mathematics and I'm stuck with question:

If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.

I've no idea how to prove this, please help me by give me some reference or hint to its solution.

Arctic Char
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idonno
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    To prove that $F^{-1}$ is an inverse of a function $F$ you need to show that $F^{-1}\circ F(x)=x$ and also $F\circ F^{-1}(x)=x$ – Leandro Aug 13 '10 at 14:39
  • Please try to use more descriptive titles when asking questions. – Akhil Mathew Aug 13 '10 at 15:11
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    @Akhil: could you suggest more descriptive title for this question please, I'm not very good in English. – idonno Aug 13 '10 at 17:49
  • @Pete: It's been edited since when I posted the comment. – Akhil Mathew Aug 13 '10 at 23:20
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    @Akhil: It has had this title (except with "proof" instead of "prove") even before your comment. I think what happened is that when you saw the question on the main page, the math was not rendered immediately, and you saw "How to proof ?". – ShreevatsaR Aug 14 '10 at 06:09
  • @ShreevatsaR: I think you're correct; actually, I had not set up the math rendering at that point, and assumed that the question was titled "How to proof." My apologies to idonno. – Akhil Mathew Aug 14 '10 at 14:38
  • None of the answers below is completely satisfying. For a better one, see for instance https://math.stackexchange.com/questions/1802618 – Anne Bauval Apr 06 '24 at 14:06

7 Answers7

46

You put your socks first and then your shoes but you take off your shoes before taking off your socks.

lhf
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    While funny, I think this as a stand-alone comment, is not quite helpful. If you briefly showed how it was related however, I think it would be a very useful answer. – BBischof Aug 13 '10 at 22:13
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    it's not just funny! Think of 'putting on socks' as 'applying the function f', taking them off as the inverse. and 'putting on shoes' as 'applying g'. –  Aug 14 '10 at 05:51
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    @muad: Yes, we understand, but for someone struggling to prove the statement in the title, the connection is probably not obvious. – ShreevatsaR Aug 14 '10 at 07:35
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    This is amazingly clear intuitively, but it doesn't constitute a mathematical proof – JacksonFitzsimmons Apr 30 '16 at 08:24
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    for me, is not intuitive, because there are not conceptos of bijectivity for put a sock – ESCM Jul 07 '20 at 02:17
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$$\begin{align} & \text{id} \\\\ =& f \circ f^{-1} \\\\ =& f \circ \text{id} \circ f^{-1} \\\\ =& f \circ (g \circ g^{-1}) \circ f^{-1} \\\\ =& f \circ g \circ g^{-1} \circ f^{-1} \\\\ =& (f \circ g) \circ (g^{-1} \circ f^{-1}) \end{align}$$

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

Later
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    This answer is not complete. It only proves that $f\circ g$ has a right inverse (i.e. is surjective), not that it is bijective, hence the final sentence is unproved. Moreover, the two $\operatorname{id}$ (which are not the same) should be specified (id on which set?). – Anne Bauval Apr 06 '24 at 14:00
10

Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.

Dylan Wilson
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  • Right hand side mean both (f o g) -1 and g-1 o f-1 ? – idonno Aug 13 '10 at 14:39
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    I think the idea of Dylan is that if you call $h=f\circ g$ then we have that $h\circ (g^{-1} \circ f^{-1}) = (f\circ g) \circ (g^{-1} \circ f^{-1}) = Id$, using the associativity of composition, which shows that $g^{-1} \circ f^{-1}$ is the inverse of $h$. – Ismael Aug 14 '10 at 00:30
  • No, it is not sufficient to prove the claim. Same flaw as in this answer (see more detailed comment below it). – Anne Bauval Apr 06 '24 at 14:14
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$f:Y \to Z$ and $g:X \to Y$ are invertible functions. We need to prove $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.

$$ (f\circ g)^{-1}\circ (f\circ g)=(g^{-1}\circ f^{-1})\circ(f\circ g)=(g^{-1}\circ f^{-1})\circ f)\circ g=(g^{-1}\circ (f^{-1}\circ f))\circ g=(g^{-1}\circ I_{Y})\circ g=g^{-1}\circ g=I_{X} $$ Similarly, $$ (f\circ g)\circ (f\circ g)^{-1}=(f\circ g)\circ (g^{-1}\circ f^{-1})=f\circ(g\circ (g^{-1}\circ f^{-1}))=f\circ((g\circ g^{-1})\circ f^{-1})\\=f\circ (I_{Y}\circ f^{-1})=(f\circ I_{Y})\circ f^{-1}=f\circ f^{-1}=I_{Z} $$

$(g^{-1}\circ f^{-1})\circ(f\circ g)=I_{X}$ and $(f\circ g)\circ (g^{-1}\circ f^{-1})=I_{Y}$ proves $f\circ g$ is invertible with $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.

SOORAJ SOMAN
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    Am I missing something? The first steps are using the equality you'd like to prove. Also, shouldn't the last line be $(f\circ g)\circ (g^{-1}\circ f^{-1})=I_{Z}$, and not $=I_{Y}$? – Eric Duminil Jan 24 '23 at 10:07
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    @EricDuminil You are right, this alledged answer is worthless. None of the other answers here being completely satisfying, better look here. – Anne Bauval Apr 06 '24 at 14:08
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let $(x,y) \in (f \circ g) ^{-1} $
$\Leftrightarrow (y,x) \in (f \circ g)$
$\Leftrightarrow \exists t((y,t) \in g \land (t,x) \in f )$
$\Leftrightarrow \exists t((t,y) \in g^{-1} \land (x,t) \in f^{-1} )$
$\Leftrightarrow (x,y) \in g^{-1} \circ f^{-1}$

Thiago Figueredo
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Mark
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Heres a hint: The jacket is put on after the shirt, but is taken off before it.

-4

This is straightforward. Take x in the domain of f. It goes to f(x) = y. And g takes y to z = g(y). Therefore $g^{-1}$ takes z to y and $f^{-1}$ takes y to x. Both sides of your equation takes $z$ to $x$.

Please try to think more before asking. This was not hard, was it? :)

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    I'm reading this book by myself, sometime it's seem very hard fro me to understand all of material without any suggestion. Anyway, I appreciate your help. =) – idonno Aug 13 '10 at 14:44
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    Sorry about the abrupt tone. I didn't mean to be rude. I have now softened it. –  Aug 13 '10 at 14:48
  • I apologize for my misunderstanding too. – idonno Aug 14 '10 at 07:38