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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.

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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$2010-08-13
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    Please try to use more descriptive titles when asking questions.2010-08-13
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    @Akhil: could you suggest more descriptive title for this question please, I'm not very good in English.2010-08-13
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    @Pete: It's been edited since when I posted the comment.2010-08-13
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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 ?".2010-08-14
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    @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.2010-08-14

8 Answers 8

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Let f:A→B and g:B→C be functions and let H be a subset of C. then we have (g∘f)^(-1) (H)=f^(-1) (g^(-1) (H)). Note the reversal in the order of the functions.

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

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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.2010-08-13
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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'.2010-08-14
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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.2010-08-14
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    This is amazingly clear intuitively, but it doesn't constitute a mathematical proof2016-04-30
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Use the definition of an inverse and associativity of composition to show that the right hand side is the inverse of $(f \circ g)$.

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    Right hand side mean both (f o g) -1 and g-1 o f-1 ?2010-08-13
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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 $h$.2010-08-14
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$$\begin{align} & \text{id} \\ =& f \circ f^\circ \\ =& f \circ \text{id} \circ f^\circ \\ =& f \circ (g \circ g^\circ) \circ f^\circ \\ =& f \circ g \circ g^\circ \circ f^\circ \\ =& (f \circ g) \circ (g^\circ \circ f^\circ) \end{align}$$

Therefore $(f \circ g)^\circ = g^\circ \circ f^\circ$.

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

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    This is essential what lhf said.2012-05-12
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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. =)2010-08-13
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    Sorry about the abrupt tone. I didn't mean to be rude. I have now softened it.2010-08-13
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    I apologize for my misunderstanding too.2010-08-14
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let $(x,y) \in (f \circ g) ^{-1} $
$\Leftrightarrow (y,x) \in (f \circ g)$
$\Leftrightarrow \exists t((y,t) \in g \land (t,x) \in f )$
$\Leftrightarrow \exists t((t,y) \in g^{-1} ) \land \exists ((x,t) \in f^{-1} ))$
$\Leftrightarrow (x,y) \in g^{-1} \circ f^{-1}$

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