Selected items from the Introduction to set theory and to methodology and philosophy of mathematics and computer programing.

1. Introduction to set theory and to methodology and philosophy of
mathematics and computer programming
Function composition
An overview
by Jan Plaza
c 2017 Jan Plaza
Use under the Creative Commons Attribution 4.0 International License
Version of April 29, 2017

2. Definition. Let f and g be functions such that range(f) ⊆ domain(g).
The function composition of f and g , denoted g
func
◦ f ,
is the unique function that has the same domain as f,
and such that (g ◦ f)(x)=g(f(x)), for every x ∈ domain(f).
Note. “g ◦ f” is read “f composed with g” or “function composition of f and g”,
as if we were reading the expression “g ◦ f” from the right to the left.
While writing “f composed with g” in symbols, write from the right to the left.
Note. This definition does not tell what g
func
◦ f is when f or g is not a function
or when range(f) domain(g), and such cases are considered not iteresting.
Proposition. Let f and g be functions such that range(f) ⊆ domain(g),
i.e. such that the function composition g ◦ f is defined.
Then, (function composition) g
func
◦ f = (relation composition) g ◦ f .
Convention. Instead of
func
◦ we will write just ◦ . The context should tell whether the
relation composition or function composition is meant.

3. If f : X1 −→ Y1 and g : X2 −→ Y2 and Y1 ⊆ X2
then function composition g ◦ f is defined, and g ◦ f : X1 −→ Y2.
f
g
g ◦ f
X1 Y1
X2 Y2

4. Equivalently, if f : X −→ Y and g : Y −→ Z
then function composition g ◦ f is defined, and g ◦ f : X −→ Z.
f
g
g ◦ f
X
Y Z

5. Example
f
g
g ◦ f
1
2
3
10
11
12
13
14
20
22
24
f = { 1, 11 , 2, 12 , 3, 12 }
g = { 10, 20 , 11, 20 , 12, 22 , 13, 24 , 14, 24 }
g ◦ f = { 1, 20 , 2, 22 , 3, 22 }

6. Example
Let f : R −→ R where f(x)=2x.
Let g : R −→ R where g(x)=x + 1.
We have (f ◦ g)(x)=f(g(x))=2(x + 1) = 2x + 2.
We have (g ◦ f)(x)=g(f(x))=2x + 1.
Notice that f ◦ g=g ◦ f.

7. Example
Let f : R −→ R be defined as f(x) = sin x.
Let g : R 0 −→ R be defined as g(x) =
√
x.
1. Function composition g ◦ f is not defined
because range(f) = [−1, 1,] contains negative numbers, not in domain(g).
2. Relation composition g ◦ f is defined and results in a partial function on R;
g ◦ f : R −→ R such that
(g ◦ f)(x) =
√
sin x for x ∈ [2kπ, (2k + 1)π] where k ∈ Z.
3. Function composition f ◦ g is defined, and
it is the same as relation composition f ◦ g.
It results in a function f ◦ g : R 0 −→ R such that
(f ◦ g)(x) = sin(
√
x).

8. Fact
Let f, g be functions such that function composition g ◦ f is defined. Then:
1. domain(g ◦ f) = domain(f)
2. range(g ◦ f) ⊆ range(g)
Exercise
Let f, g be functions such that function composition g ◦ f is defined.
1. Disprove: domain(g ◦ f) = domain(g)
2. Disprove: range(g ◦ f) = range(g)