Inverses

A function may or may not have an inverse. The obstruction is already in Definition 1.1: each input has exactly one output, but two inputs may share an output. If they do, I cannot run the rule backwards.

Definition 1.2injective

𝑓:𝐴\to 𝐵

is injective if

𝑓({𝑎}_{1})=𝑓({𝑎}_{2})

implies

{𝑎}_{1}={𝑎}_{2}

Definition 1.3inverse

𝑓:𝐴\to 𝐵

is injective, the inverse

{𝑓}^{-1}:𝑓(𝐴)\to 𝐴

is the function that sends

𝑓(𝑎)

back to

𝑎

. Equivalently,

{𝑓}^{-1}\circ 𝑓={𝐼}_{𝐴}

and

𝑓\circ {𝑓}^{-1}={𝐼}_{𝑓(𝐴)}

The second line uses Proposition 1.1. I have to write the codomain of ${𝑓}^{-1}$ as the image $𝑓(𝐴)$ , not as $𝐵$ , unless $𝑓$ hits everything. Surjectivity onto $𝐵$ is a separate request; injectivity is what makes a left inverse exist.

Theorem 1.2

An injective function has at most one inverse.

Proof

Suppose

𝑔

and

h

both invert

𝑓:𝐴\to 𝐵

𝑓(𝐴)

. For

𝑏=𝑓(𝑎)

in the image,

𝑔(𝑏)=𝑔(𝑓(𝑎))=𝑎=h(𝑓(𝑎))=h(𝑏)

. So

𝑔=h

as functions

𝑓(𝐴)\to 𝐴

Cutting the domain until it works

Absolute value on $\mathbb{R}$ is not injective: $|-2|=|2|$ . There is no inverse on the whole line. Restrict to $[0,\infty )$ and $𝑥\mapsto |𝑥|$ becomes the identity, which is its own inverse. Restrict to $(-\infty ,0]$ and the inverse is $𝑥\mapsto -𝑥$ . Same formula, two domains, two different invertible functions. That is Definition 1.1 again: the domain is part of the data.

Example 1.3

𝑠(𝑥)={𝑥}^{2}

\mathbb{R}

is not injective. On

[0,\infty )

it is, and the inverse is

𝑥\mapsto \sqrt{𝑥}

. On

(-\infty ,0]

the inverse is

𝑥\mapsto -\sqrt{𝑥}

. Compare Example 1.2 and Composition: after the restriction,

𝑟\circ 𝑠

and

𝑠\circ 𝑟

become identities on the cut domains.

Inverse of a composite

If I can invert each factor, I can invert the composite, but the order flips.

Proposition 1.2

𝑓:𝐴\to 𝐵

and

𝑔:𝐵\to 𝐶

are injective, then

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

on the image of

𝑔\circ 𝑓

Proof

({𝑓}^{-1}\circ {𝑔}^{-1})\circ (𝑔\circ 𝑓)={𝑓}^{-1}\circ ({𝑔}^{-1}\circ 𝑔)\circ 𝑓={𝑓}^{-1}\circ 𝑓={𝐼}_{𝐴}

by (1.4) and Definition 1.3. The other composite is the identity on the image by the same parentheses.

I will use this later without thinking: the chain rule differentiates a composite in the same reversed order that inversion undoes it.

Exercise 1.3

Show that

𝑓(𝑥)={𝑥}^{3}-𝑥

\mathbb{R}

is not injective. Find three points with the same image, or argue from the derivative after you have it.

Solution to Exercise 1.3

𝑓(-1)=0=𝑓(0)=𝑓(1)

. Three distinct inputs, one output, so Definition 1.2 fails. After (2.1) one can also note that

{𝑓}^{\prime }

changes sign, but three roots of

𝑓(𝑥)=0

are enough.