inverse of composition of functions

Theorem.

Let $f$ and $g$ be invertible functions such that their composition $f\mathrm{\circ }g$ is well defined. Then $f\mathrm{\circ }g$ is invertible and

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

Before proving this theorem, it should be noted that some students encounter this result long before they are introduced to formal proof. Fortunately, there is an intuitive way to think about this theorem: Think of the function $g$ as putting on one’s socks and the function $f$ as putting on one’s shoes. Then $f\circ g$ denotes the process of putting one one’s socks, then putting on one’s shoes. (Recall that function composition works from right to left.) Note that ${(f\circ g)}^{-1}$ refers to the reverse process of $f\circ g$, which is taking off one’s shoes (which is $f^{-1}$) followed by taking off one’s socks (which is $g^{-1}$).

Due to the intuitive argument given above, the theorem is referred to as the socks and shoes rule. This name is a mnemonic device which reminds people that, in order to obtain the inverse of a composition of functions, the original functions have to be undone in the opposite order.

Now for the formal proof.

Proof.

Let $A$, $B$, and $C$ be sets such that $g:A\to B$ and $f:B\to C$. Then the following two equations must be shown to hold:

$(g^{-1}\circ f^{-1})\circ (f\circ g)={\text{id}}_A$ (1)

$(f\circ g)\circ (g^{-1}\circ f^{-1})={\text{id}}_C$ (2)

Note that ${\text{id}}_X$ denotes the identity function on the set $X$.

The two equations given above follow easily from the fact that function composition is associative.

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

$=g^{-1}\circ ({\text{id}}_B\circ g)$

$=g^{-1}\circ g$

$={\text{id}}_A$

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

$=f\circ ({\text{id}}_B\circ f^{-1})$

$=f\circ f^{-1}$

$={\text{id}}_C\mathit{∎}$

The socks and shoes rule has a natural generalization:

Corollary.

Let $n$ be a positive integer and $f_{\mathrm{1}}\mathrm{,}\mathrm{\dots }\mathrm{,}{f}_n$ be invertible functions such that their composition $f_{\mathrm{1}}\mathrm{\circ }\mathrm{\dots }\mathrm{\circ }{f}_n$ is well defined. Then $f_{\mathrm{1}}\mathrm{\circ }\mathrm{\dots }\mathrm{\circ }{f}_n$ is invertible and

$${(f_1\circ \mathrm{\dots }\circ f_n)}^{-1}=f_n{}^{-1}\circ \mathrm{\dots }\circ f_1{}^{-1}.$$

A sketch of a proof is as follows: Using induction on $n$, the socks and shoes rule can be applied with $f=f_1\circ \mathrm{\dots }\circ f_{n-1}$ and $g=f_n$.

Title	inverse of composition of functions
Canonical name	InverseOfCompositionOfFunctions
Date of creation	2013-03-22 17:47:47
Last modified on	2013-03-22 17:47:47
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	8
Author	Wkbj79 (1863)
Entry type	Theorem
Classification	msc 03-00
Classification	msc 03E20
Classification	msc 97D40
Synonym	socks and shoes rule
Related topic	Function
Related topic	InverseFormingInProportionToGroupOperation