10
$\begingroup$

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.

  • 0
    @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

-2

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.

31

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

  • 2
    This is amazingly clear intuitively, but it doesn't constitute a mathematical proof2016-04-30
10

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

  • 0
    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
10

$\begin{align} & \text{id} \\ =& f \circ f^\circ \\ =& f \circ \text{id} \circ f^\circ \\ =& f \circ (g \circ g^\circ) \circ f^\circ \\ =& f \circ g \circ g^\circ \circ f^\circ \\ =& (f \circ g) \circ (g^\circ \circ f^\circ) \end{align}$

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

3

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

  • 2
    This is essential what lhf said.2012-05-12
1

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? :)

  • 0
    I apologize for my misunderstanding too.2010-08-14
1

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

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