f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) (2) f 1
2 (3) f(2.5)
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2 (3) f(2.5)
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5)
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.
Solution To draw the graph of f, divide the domain [−2,6] into three subintervals: [−2,0), [0,2) and [2,6].
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.
Solution To draw the graph of f, divide the domain [−2,6] into three subintervals: [−2,0), [0,2) and [2,6].
-2 2 4 6
-2 -1 1 2 3 4
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.
Solution To draw the graph of f, divide the domain [−2,6] into three subintervals: [−2,0), [0,2) and [2,6].
-2 2 4 6
-2 -1 1 2 3 4
f(x) =
x2 if − 2 ≤ x < 0
2x if 0 ≤ x < 2
4 − x if 2 ≤ x ≤ 6
Find the value of
(1) f(−1) = (−1)2 = 1 (2) f 1
2
= 2 · 1
2 = 1
(3) f(2.5) = 4 − 2.5 = 1.5 and sketch the graph of f.
Solution To draw the graph of f, divide the domain [−2,6] into three subintervals: [−2,0), [0,2) and [2,6].
-2 2 4 6
-2 -1 1 2 3 4
|x| =
x if x ≥ 0
|x| =
x if x ≥ 0
−x if x < 0 Example
|x| =
x if x ≥ 0
−x if x < 0 Example
(1) |2| = 2
|x| =
x if x ≥ 0
−x if x < 0 Example
(1) |2| = 2
|x| =
x if x ≥ 0
−x if x < 0 Example
(1) |2| = 2
(2) |−3| = −(−3) = 3
|x| =
x if x ≥ 0
−x if x < 0 Example
(1) |2| = 2
(2) |−3| = −(−3) = 3
• |a| is the distance from a to 0.
> | −3 | 0 | 2 3units
|x| =
x if x ≥ 0
−x if x < 0 Example
(1) |2| = 2
(2) |−3| = −(−3) = 3
• |a| is the distance from a to 0.
> | −3 | 0 | 2 3units
z }| {z }|2units {
> |
−a
| 0
| a z }| {z }| {
> |
−a
| 0
| a z }| {z }| {
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
= 5 = distance from −3 to 2
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
= 5 = distance from −3 to 2
• √a2 _{=} _{|a|}
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
= 5 = distance from −3 to 2
• √a2 _{=} _{|a|}
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
= 5 = distance from −3 to 2
• √a2 _{=} _{|a|}
Example √32 _{=} √_{9} _{=} _{3} _{=} _{|}_{3}_{|} p
> |
−a
| 0
| a z }| {z }| {
• |a − b| is the distance from a to b. Example a = −3, b = 2
|−3 − 2| = | − 5|
= 5 = distance from −3 to 2
• √a2 _{=} _{|a|}
Example √32 _{=} √_{9} _{=} _{3} _{=} _{|}_{3}_{|} p
Graph of the absolute value function f(x) = |x|
Divide the real number line into two subintervals [0,∞) and (−∞,0)
f(x) =
x if x ≥ 0 −x if x < 0
Graph of the absolute value function f(x) = |x|
Divide the real number line into two subintervals [0,∞) and (−∞,0)
f(x) =
x if x ≥ 0 −x if x < 0
-2 -1 1 2
0.5 1 1.5
Graph of the absolute value function f(x) = |x|
Divide the real number line into two subintervals [0,∞) and (−∞,0)
f(x) =
x if x ≥ 0 −x if x < 0
-2 -1 1 2
0.5 1 1.5
y =
1 − x if x ≥ 0
y =
1 − x if x ≥ 0
1 − (−x) if x < 0
-2 -1 1 2
-1 -0.5
0.5 1
y =
1 − x if x ≥ 0
1 − (−x) if x < 0
-2 -1 1 2
-1 -0.5
0.5 1
y =
1 − x if x ≥ 0
1 − (−x) if x < 0
-2 -1 1 2
-1 -0.5
0.5 1
y =
1 − x if x ≥ 0
1 − (−x) if x < 0
-2 -1 1 2
-1 -0.5
0.5 1
y =
1 − x if x ≥ 0
1 − (−x) if x < 0
-2 -1 1 2
-1 -0.5
0.5 1
Example Graph y = |x − 1|
y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
Example Graph y = |x − 1|
y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
-1 1 2 3
0.5 1 1.5
Example Graph y = |x − 1|
y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
-1 1 2 3
0.5 1 1.5
Example Graph y = |x − 1| y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
-1 1 2 3
0.5 1 1.5
2
-2 -1 1 2 3
0.5 1 1.5
Example Graph y = |x − 1| y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
-1 1 2 3
0.5 1 1.5
2
-2 -1 1 2 3
0.5 1 1.5
Example Graph y = |x − 1| y =
x − 1 if x − 1 ≥ 0 −(x − 1) if x − 1 < 0
-1 1 2 3
0.5 1 1.5
2
-2 -1 1 2 3
0.5 1 1.5
• First, square both sides to get y2 = x.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
• First, square both sides to get y2 = x. Its graph is a parabola.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
√
x is always non-negative.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
√
x is always non-negative.
Graph of y = √x is the upper half of the parabola.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
x = y2
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
√
x is always non-negative.
Graph of y = √x is the upper half of the parabola.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
x = y2
y = √x
Note y = √x
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
√
x is always non-negative.
Graph of y = √x is the upper half of the parabola.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
x = y2
y = √x
Note y = √x
y2 = x, y ≥ 0 Reason a = b =⇒ a2 = b2
• First, square both sides to get y2 = x. Its graph is a parabola.
• Squarring introduces extra points.
√
x is always non-negative.
Graph of y = √x is the upper half of the parabola.
-2 -1 1 2 3 4
-2 -1
1 2 3
4 y = x 2
x = y2
y = √x
Note y = √x
y2 = x, y ≥ 0 Reason a = b =⇒ a2 = b2
(1) y = √x − 2 Solution
(1) y = √x − 2 Solution
(1) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units down.
2.5 5 7.5 10 12.5 15
-2 -1
1 2 3
4 y =
√
(1) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units down.
2.5 5 7.5 10 12.5 15
-2 -1
1 2 3
4 y =
√
(1) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units down.
2.5 5 7.5 10 12.5 15
-2 -1
1 2 3
4 y =
√
x
(2) y = √x − 2 Solution
(2) y = √x − 2 Solution
(2) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units to the right.
2 4 9 16
1 2 3 4
(2) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units to the right.
2 4 9 16
1 2 3 4
(2) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units to the right.
2 4 9 16
1 2 3 4
y = √x
(2) y = √x − 2 Solution
The graph is obtained by moving that of y = √x two units to the right.
2 4 9 16
1 2 3 4
y = √x
y = √x − 2
√
Eg. we may combine the square function and sine function to get sin(x2)
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by
(g ◦ f)(x) = g f(x)
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by
(g ◦ f)(x) = g f(x) Example Let f and g be functions given by
f(x) = x2 g(x) = sin x
Then g ◦ f : R −→ R is
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by
(g ◦ f)(x) = g f(x) Example Let f and g be functions given by
f(x) = x2 g(x) = sin x
Then g ◦ f : R −→ R is (g ◦ f)(x) = g(x2)
Eg. we may combine the square function and sine function to get sin(x2) x 7→ x2 7→ sin(x2)
Definition Let f and g be functions. The composition of g with f, denoted by g ◦ f, is the function defined by
(g ◦ f)(x) = g f(x) Example Let f and g be functions given by
f(x) = x2 g(x) = sin x
Then g ◦ f : R −→ R is (g ◦ f)(x) = g(x2)
= sin(x2)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9)
(2) (f ◦ g)(6)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
(2) (f ◦ g)(6)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3)
(2) (f ◦ g)(6)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1 (2) (f ◦ g)(6)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4) = 2 (3) (f ◦ g)(1)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4) = 2
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4) = 2
(3) (f ◦ g)(1) = f g(1)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4) = 2
(3) (f ◦ g)(1) = f g(1)
Example Let f(x) = √x and g(x) = x − 2. Find the following (if defined)
(1) (g ◦ f)(9) = g f(9)
= g(3) = 1
(2) (f ◦ g)(6) = f g(6)
= f(4) = 2
(3) (f ◦ g)(1) = f g(1)
= f(−1) undefined
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x)
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1)
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1
(2) (g ◦ f)(x) = g(f(x)) = g(x2)
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1
(2) (g ◦ f)(x) = g(f(x)) = g(x2) = 2x2 + 1
Example Let f(x) = x2 and g(x) = 2x + 1. Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x) Solution
(1) (f ◦ g)(x) = f(g(x)) = f(2x + 1) = (2x + 1)2 = 4x2 + 4x + 1
(2) (g ◦ f)(x) = g(f(x)) = g(x2) = 2x2 + 1
f(x) = 2x + 1, g(x) = x − 1 2 . Find
(1) (f ◦ g)(x) (2) (g ◦ f)(x)
f(x) = 2x + 1, g(x) = x − 1 2 . Find
(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x
f(x) = 2x + 1, g(x) = x − 1 2 . Find
(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x
Definition A function ϕ from a set X into itself satisfying ϕ(x) = x for all x ∈ X is called the identity function on X.
f(x) = 2x + 1, g(x) = x − 1 2 . Find
(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x
Definition A function ϕ from a set X into itself satisfying ϕ(x) = x for all x ∈ X is called the identity function on X.
Remark The above example means that (f ◦ g) and (g ◦ f) are the identity functions on R.
f(x) = 2x + 1, g(x) = x − 1 2 . Find
(1) (f ◦ g)(x) = x (2) (g ◦ f)(x) = x
Definition A function ϕ from a set X into itself satisfying ϕ(x) = x for all x ∈ X is called the identity function on X.
Remark The above example means that (f ◦ g) and (g ◦ f) are the identity functions on R.
Question Given a function f : X −→ Y, when can we find a function g : Y −→ X
1 2 3 4 5 6 7 A
B C D E
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C
Then (g ◦ f)(A) = A
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D
Then (g ◦ f)(A) = A
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D
Then (g ◦ f)(A) = A
(g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A
Then (g ◦ f)(A) = A
(g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B
Then (g ◦ f)(A) = A
(g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B
Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A
Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A g(7) = E
Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Yes. Define g(1) = A g(2) = C g(3) = D g(4) = A g(5) = B g(6) = A g(7) = E
Then (g ◦ f)(A) = A (g ◦ f)(B) = B (g ◦ f)(C) = C (g ◦ f)(D) = D (g ◦ f)(E) = E
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
(g ◦ f)(D) = D (g ◦ f)(E) = E
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
(g ◦ f)(D) = D (g ◦ f)(E) = E
1 2 3 4 5 6 7 A
B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
(g ◦ f)(D) = D (g ◦ f)(E) = E
In particular, g(1) = A g(1) = C
1 2 3 4 5 6 7 A B C D E
Is there any g : Y −→ X satisfying (g ◦ f)(x) = x for all x ∈ X ?
Solution Need (g ◦ f)(A) = A
(g ◦ f)(B) = B (g ◦ f)(C) = C
(g ◦ f)(D) = D (g ◦ f)(E) = E
In particular, g(1) = A g(1) = C
x1 , x2 =⇒ f(x1) , f(x2) Example
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x_{1} , x_{2} =⇒ 2x1 , 2x2
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x_{1} , x_{2} =⇒ 2x1 , 2x2
g(x) = x2 is NOT injective
-2 -1 1 2
1 2 3 4
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x_{1} , x_{2} =⇒ 2x1 , 2x2
g(x) = x2 is NOT injective
-2 -1 1 2
1 2 3 4
x1 , x2 =⇒ f(x1) , f(x2) Example
f(x) = 2x is injective
-2 -1 1 2
1 2 3 4
x_{1} , x_{2} =⇒ 2x1 , 2x2
g(x) = x2 is NOT injective
-2 -1 1 2
1 2 3 4
−1 , 1 but (−1)2 = 12
Remark Geometrically, f is injective means that graph of f intersects every horizontal line in at most one point.