How to prove $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$? (inverse of composition)

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.

• 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
• @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
• @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 6:09

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

• Right hand side mean both (f o g) -1 and g-1 o f-1 ? – idonno Aug 13 '10 at 14:39
• 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 0:30

• 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
• 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'. – anon Aug 14 '10 at 5:51
• @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 7:35
• This is amazingly clear intuitively, but it doesn't constitute a mathematical proof – JacksonFitzsimmons Apr 30 '16 at 8:24

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

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

• This is essential what lhf said. – Pedro Tamaroff May 12 '12 at 16:47

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? :)

• 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
• Sorry about the abrupt tone. I didn't mean to be rude. I have now softened it. – user1119 Aug 13 '10 at 14:48
• I apologize for my misunderstanding too. – idonno Aug 14 '10 at 7:38

let $(x,y) \in (f \circ g) ^{-1}$
$\Leftrightarrow (y,x) \in (f \circ g)$
$\Leftrightarrow \exists t((y,t) \in g \land (t,x) \in f )$
$\Leftrightarrow \exists t((t,y) \in g^{-1} ) \land \exists ((x,t) \in f^{-1} ))$
$\Leftrightarrow (x,y) \in g^{-1} \circ f^{-1}$

$f:Y \to Z$ 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 $(f\circ g)^{-1}=g^{-1}\circ f^{-1}$.

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.