Mathematical-Reasoning · Unit 23 · Video 5 · Interactive Practice

What an Injective or Surjective Composite Reveals

IKey Formulas

Statement	Name	Why it holds
$f, g$ injections $\Rightarrow g \circ f$ injection; the same for surjections and for bijections	Composition keeps the type	From $g(f(x)) = g(f(y))$ , strip $g$ (it is an injection), then strip $f$
$g \circ f$ injection $\Rightarrow f$ injection	The inner function is forced	Apply $g$ to $f(x) = f(y)$ ; every function allows that
$g \circ f$ surjection $\Rightarrow g$ surjection	The outer function is forced	For $c \in C$ pick $x$ with $g(f(x)) = c$ and set $b = f(x)$
$A = \{1\}$ , $B = \{p, q\}$ , $C = \{s\}$ , $f(1) = p$ , $g(p) = g(q) = s$	Counterexample	$g \circ f$ is a bijection, yet $f$ is not a surjection and $g$ is not an injection

Key Insight: The composite only ever applies $g$ to outputs of $f$ . An element of $B$ that $f$ misses, like $q$ , is invisible to $g \circ f$ , so whatever $g$ does there cannot be detected.

IIVisualization 1 — What Does the Composite Force?

Which properties of $f$ and $g$ does a given composite $g \circ f$ actually force, and which can fail?

Sizes of A, B, C

	$f$ (inner)	$g$ (outer)	$g \circ f$
injection
surjection

💡 When $|A| = |B|$ is finite, an injective $f$ reaches all of $B$ , so no element like $q$ is left over, and an injective $g \circ f$ then forces $g$ to be an injection as well.

IIIVisualization 2 — The Proof That $g$ Is a Surjection

Each $c \in C$ gets its preimage $b = f(x)$ under $g$ from the composite, with no need for $f$ to be onto.

Element c Proof step 1

IVQuiz Questions

Problem 1 · What an Injective Composite Guarantees

Given: nonempty sets $A$ , $B$ , $C$ and functions $f : A \to B$ , $g : B \to C$ such that $g \circ f$ is an injection. Which statement must be true?

✅ Correct! An injective composite forces the inner function

f

to be an injection, and nothing more.

❌ Not quite. In the example

A = \{1\}

B = \{p, q\}

C = \{s\}

f(1) = p

g(p) = g(q) = s

, the composite is an injection but

g(p) = g(q)

with

p \ne q

❌ Not quite. Injectivity of

g \circ f

says nothing about reaching all of

B

: in the example

f(1) = p

, nothing goes to

q

, yet

g \circ f

is an injection.

❌ This is the common guess. It claims too much:

f

must be an injection, but the example with

g(p) = g(q) = s

has an injective composite and a non-injective

g

Show solution

Let $x, y \in A$ with $f(x) = f(y)$ . Since $g$ is a function, applying it to equal inputs gives equal outputs:

g(f(x)) = g(f(y)), \quad\text{that is,}\quad (g \circ f)(x) = (g \circ f)(y).

Since $g \circ f$ is an injection, $x = y$ . So $f$ is an injection. The proof used nothing about $g$ .

The other options fail for $A = \{1\}$ , $B = \{p, q\}$ , $C = \{s\}$ , $f(1) = p$ , $g(p) = g(q) = s$ : here $g \circ f$ is an injection, $g$ is not an injection, and $f$ is not a surjection.

Problem 2 · A Collision the Composite Never Sees

Given: $A = \{1, 2\}$ , $B = \{p, q, r\}$ , $C = \{s, t\}$ , with $f(1) = q$ , $f(2) = r$ and $g(p) = s$ , $g(q) = s$ , $g(r) = t$ . Which statement is true?

✅ Correct!

(g \circ f)(1) = s

and

(g \circ f)(2) = t

, so the composite is a bijection, while

g(p) = g(q)

happens at

p

, which

f

never reaches.

❌ Not quite. The composite applies

g

only to

f(1) = q

and

f(2) = r

. Their images

s

and

t

are different, so the collision at

p

never enters

g \circ f

❌ Wrong function. A surjective composite forces the outer function

g

to be a surjection. Here

f

misses

p

❌ Wrong function. An injective composite forces the inner function

f

to be an injection. Here

g(p) = g(q) = s

with

p \ne q

Show solution

Compute the composite: $(g \circ f)(1) = g(q) = s$ and $(g \circ f)(2) = g(r) = t$ . Its two inputs have different outputs, and both elements of $C$ are reached, so $g \circ f : A \to C$ is a bijection.

The parts: $f$ is an injection ( $q \ne r$ ) but not a surjection (nothing goes to $p$ ); $g$ is a surjection but not an injection ( $g(p) = g(q) = s$ ). Both failures happen at $p$ , the element of $B$ outside the range of $f$ , which is why the composite cannot detect them.

Problem 3 · Compose, Then Diagnose

Given: $A = \{1, 2, 3\}$ , $B = \{p, q, r, s\}$ , $C = \{u, v, w\}$ , with $f(1) = q$ , $f(2) = s$ , $f(3) = p$ and $g(p) = w$ , $g(q) = u$ , $g(r) = u$ , $g(s) = v$ .

What is $(g \circ f)(3)$ ?

Which description is correct?

✅ Correct!

g \circ f

sends

1 \mapsto u

2 \mapsto v

3 \mapsto w

, and as the theorem promises,

f

is an injection and

g

is a surjection.

❌ Follow both arrows. First

f(3) = p

, then

g(p) = w

❌ That is only halfway.

f(3) = p

is an element of

B

; the composite continues with

g(p) = w \in C

❌ The converse fails.

f

misses

r

, so it is not a surjection, and

g(q) = g(r)

, so

g

is not an injection.

❌ Not quite. The composite applies

g

only to

f(1) = q

f(2) = s

f(3) = p

; the element

r

is never an output of

f

, so

g(r) = u

never enters

g \circ f

❌ Swapped.

f

has three different outputs

q, s, p

but misses

r

;

g

reaches

u, v, w

but sends

q

and

r

to the same

u

Show solution

$(g \circ f)(1) = g(q) = u, \quad (g \circ f)(2) = g(s) = v, \quad (g \circ f)(3) = g(p) = w.$

The three outputs are different and cover $C$ , so $g \circ f : A \to C$ is a bijection.

$f$ : its outputs $q, s, p$ are different, so $f$ is an injection; nothing goes to $r$ , so $f$ is not a surjection.

$g$ : it reaches $u, v, w$ , so it is a surjection; $g(q) = g(r) = u$ with $q \ne r$ , so it is not an injection.

This matches the theorem: an injective composite forces $f$ injective, a surjective composite forces $g$ surjective, and the other two properties fail, both at $r$ .

Problem 4 · An Infinite Version of the Counterexample

Given: $T = \{x \in \mathbb{R} \mid x \ge 0\}$ , $f : T \to \mathbb{R}$ with $f(x) = x$ , and $g : \mathbb{R} \to T$ with $g(x) = x^2$ . Then $g \circ f : T \to T$ . Which statement is true?

✅ Correct! Both failures happen at negative numbers, which

f

never outputs, so the composite

x \mapsto x^2

T

cannot see them.

❌ Not quite. The composite applies

g

only to outputs of

f

, which are all

\ge 0

; the input

-1

is never one of them. For

a, b \in T

a^2 = b^2

gives

a = b

❌ The converse fails.

-1 \in \mathbb{R}

is not

f(x)

for any

x \in T

, and

g(-1) = g(1) = 1

with

-1 \ne 1

❌ Check the codomain. The codomain of

g \circ f

T

, which has no negative numbers to miss; every

c \in T

equals

(g \circ f)(\sqrt{c})

Show solution

For $x \in T$ , $(g \circ f)(x) = g(x) = x^2$ .

Injection: if $a, b \in T$ and $a^2 = b^2$ , then $(a - b)(a + b) = 0$ . If $a + b = 0$ then $a = b = 0$ , because both are $\ge 0$ ; otherwise $a = b$ .

Surjection: for $c \in T$ , the number $\sqrt{c} \in T$ satisfies $(g \circ f)(\sqrt{c}) = c$ .

So $g \circ f$ is a bijection, and by the theorem $f$ is an injection and $g$ is a surjection. But $f$ is not a surjection ( $-1$ is never an output) and $g$ is not an injection ( $g(-1) = g(1) = 1$ ). Both failures sit at negative numbers, outside the range of $f$ .

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What an Injective or Surjective Composite Reveals

IKey Formulas

IIVisualization 1 — What Does the Composite Force?

IIIVisualization 2 — The Proof That ggg Is a Surjection

IVQuiz Questions

IIIVisualization 2 — The Proof That $g$ Is a Surjection