Composition

Once I have two functions I can feed the output of one into the other, provided the types match. If $𝑓:𝐴\to 𝐵$ and $𝑔:𝐵\to 𝐶$ , the composition $𝑔\circ 𝑓$ is the function $𝐴\to 𝐶$ given by

(𝑔\circ 𝑓)(𝑥)=𝑔(𝑓(𝑥)).

(1.3)

Theorem 1.1

𝑓:𝐴\to 𝐵

and

𝑔:𝐵\to 𝐶

are functions, then

𝑔\circ 𝑓:𝐴\to 𝐶

is a function.

Proof

Each

𝑎\in 𝐴

has exactly one

𝑓(𝑎)\in 𝐵

by Definition 1.1, and that value has exactly one

𝑔(𝑓(𝑎))\in 𝐶

. The composite assigns exactly one output to each input, which is (1.3).

The graph of a composite is not the composite of the graphs; compare (1.1). Pairing $(𝑥,𝑓(𝑥))$ with $(𝑓(𝑥),𝑔(𝑓(𝑥)))$ and forgetting the middle coordinate is a different operation from composing two subsets of the plane.

The order is the order of application, read right to left: $𝑓$ first, then $𝑔$ . I keep forgetting this and then a computation reminds me.

Algebra of composition

Composition is associative whenever it is defined,

h\circ (𝑔\circ 𝑓)=(h\circ 𝑔)\circ 𝑓,

(1.4)

because both sides send $𝑥$ to $h(𝑔(𝑓(𝑥)))$ . It is not commutative. Take $𝑓(𝑥)=𝑥+1$ and $𝑔(𝑥)=2𝑥$ . Then

(𝑔\circ 𝑓)(𝑥)=2𝑥+2,\quad (𝑓\circ 𝑔)(𝑥)=2𝑥+1.

Same two functions, two different composites. The identity is the two-sided unit: $𝑓\circ {𝐼}_{𝐴}=𝑓={𝐼}_{𝐵}\circ 𝑓$ when $𝑓:𝐴\to 𝐵$ .

Proposition 1.1

𝑓:𝐴\to 𝐵

, then

𝑓\circ {𝐼}_{𝐴}=𝑓

and

{𝐼}_{𝐵}\circ 𝑓=𝑓

Proof

(𝑓\circ {𝐼}_{𝐴})(𝑎)=𝑓({𝐼}_{𝐴}(𝑎))=𝑓(𝑎)

and

({𝐼}_{𝐵}\circ 𝑓)(𝑎)={𝐼}_{𝐵}(𝑓(𝑎))=𝑓(𝑎)

When composition is defined

The rule is strict. $𝑔\circ 𝑓$ exists, with domain $𝐴$ , only if every value of $𝑓$ lands in the domain of $𝑔$ . If $𝑓:\mathbb{R}\to \mathbb{R}$ is $𝑓(𝑥)={𝑥}^{2}$ and $𝑔:(0,\infty )\to \mathbb{R}$ is $𝑔(𝑥)=\frac{1}{𝑥}$ , then $𝑔\circ 𝑓$ is not defined on all of $\mathbb{R}$ : $𝑓(0)=0$ is not in $(0,\infty )$ . Restrict the domain of $𝑓$ to $\mathbb{R}∖\{0\}$ and the composite exists,

(𝑔\circ 𝑓)(𝑥)=\frac{1}{{𝑥}^{2}}.

I would rather notice the missing point than invent a value at zero.

Example 1.2

Let

𝑠(𝑥)={𝑥}^{2}

\mathbb{R}

and

𝑟(𝑥)=\sqrt{𝑥}

[0,\infty )

. Then

𝑟\circ 𝑠

is not defined on

\mathbb{R}

, because

𝑠

takes negative values. Restrict

𝑠

[0,\infty )

and

𝑟\circ 𝑠

is the identity there. The other way around,

𝑠\circ 𝑟

is defined on

[0,\infty )

, and

(𝑠\circ 𝑟)(𝑥)=𝑥

. So one order is a left inverse on a smaller domain, and the other is not even typed until I cut.

Reading a long chain

A long expression like $\sin ({𝑒}^{{𝑥}^{2}})$ is a composition written inside out. Unpacking it,

𝑥\mapsto {𝑥}^{2}\mapsto {𝑒}^{{𝑥}^{2}}\mapsto \sin ({𝑒}^{{𝑥}^{2}}).

(1.5)

When I later differentiate, this is the chain I will walk backwards. Associativity (1.4) says I may parenthesize the chain however I like; the derivative will care about the order of the factors, not about where I put the parentheses.

Exercise 1.2

Let

𝑔(𝑥)=|𝑥|

and

𝑓(𝑥)=𝑥-1

. Compute

(𝑔\circ 𝑓)(𝑥)

and

(𝑓\circ 𝑔)(𝑥)

. Where do they differ?

Solution to Exercise 1.2

(𝑔\circ 𝑓)(𝑥)=|𝑥-1|

and

(𝑓\circ 𝑔)(𝑥)=|𝑥|-1

. They agree for

𝑥\geq 1

; at

𝑥=0

one is

1

and the other is

-1