Mathematical-Reasoning · Unit 23 · Video 4 · Interactive Practice

Composing Surjections: Pull c Back Through g, Then f

IKey Formulas

Formula	Name	Meaning
$(g \circ f)(x) = g(f(x))$	The composite	For $f\colon A \to B$ and $g\colon B \to C$ , the composite $g \circ f\colon A \to C$ applies $f$ first, then $g$
$\forall c \in C,\; \exists x \in A,\; (g \circ f)(x) = c$	$g \circ f$ is a surjection	The goal; it begins with $\forall$ , so the proof chooses an arbitrary $c \in C$
$g(b) = c$ and $f(x) = b$	The two pull-backs	$g$ onto $C$ gives some $b \in B$ ; then $f$ onto $B$ gives some $x \in A$
$(g \circ f)(x) = g(f(x)) = g(b) = c$	The closing chain	The input $x$ found by the pull-backs lands on $c$

Key Insight: The composite applies $f$ first and $g$ second, but the proof searches in the opposite order: from $c$ back through $g$ to the middle set, then back through $f$ to the domain.

IIVisualization 1 — Three Composites, Two Arrows at a Time

In each of the video's three examples, $g \circ f$ keeps the property that $f$ and $g$ share.

Example Trace

IIIVisualization 2 — The Proof Runs Backward

To reach an arbitrary $c$ , the proof pulls it back through $g$ to some $b$ , then through $f$ to some $x$ .

Step 1

1 · Hypotheses and goal

Assume $f\colon A \to B$ and $g\colon B \to C$ are surjections. Goal: $\forall c \in C,\ \exists x \in A$ with $(g \circ f)(x) = c$ .

2 · Choose an element

Let $c \in C$ be arbitrary. New goal: find $x \in A$ with $(g \circ f)(x) = c$ .

3 · Pull back through $g$

Since $g$ is a surjection, there exists $b \in B$ such that $g(b) = c$ .

4 · Pull back through $f$

Since $f$ is a surjection, there exists $x \in A$ such that $f(x) = b$ .

5 · Evaluate the composite

(g \circ f)(x) = g(f(x)) = g(b) = c

6 · Conclude

Since $c$ was arbitrary, $\forall c \in C,\ \exists x \in A$ with $(g \circ f)(x) = c$ , so $g \circ f$ is a surjection. $\blacksquare$

IVVisualization 3 — Each Pull-Back Uses One Surjection

The step into $B$ needs an element with $g(b) = c$ ; the step into $A$ needs one with $f(x) = b$ .

f(a) f(b) f(c) g(p) g(q) g(r) Pull back

💡 Challenge: make $f$ miss an element of $B$ while $g \circ f$ is still onto $C$ . The theorem's hypotheses are enough to make $g \circ f$ onto, but $f$ being onto is not necessary for it.

VQuiz Questions

Problem 1 · Pull Back on the Real Line

Given: the surjections $f\colon \mathbb{R} \to \mathbb{R}$ , $f(x) = x - 3$ , and $g\colon \mathbb{R} \to \mathbb{R}$ , $g(y) = 5y$ . Following the proof that $g \circ f$ is onto, with $c = 10$ , which middle element $b$ and input $x$ does it produce?

✅ Correct!

g(2) = 10

and

f(5) = 2

, so

(g \circ f)(5) = g(2) = 10

❌ That pulls back through $f$ first. Solving

f(b) = 10

and then

g(x) = 13

inverts

f \circ g

, not

g \circ f

: here

(g \circ f)(\tfrac{13}{5}) = 5(\tfrac{13}{5} - 3) = -2

. The outer function

g

is undone first.

❌ That runs forward from $c$ . Computing

g(10) = 50

and

f(50) = 47

applies the functions to

c

instead of finding an input that lands on

c

; in fact

(g \circ f)(47) = 220

❌ Check the second pull-back.

b = 2

is right, but

f(x) = b

means

x - 3 = 2

, so

x = b + 3

, not

b - 3

❌ Not quite. First solve

g(b) = 10

, then solve

f(x) = b

Show solution

Pull back through $g$ : we need $b \in \mathbb{R}$ with $g(b) = 10$ , that is $5b = 10$ , so $b = 2$ .

Pull back through $f$ : we need $x \in \mathbb{R}$ with $f(x) = 2$ , that is $x - 3 = 2$ , so $x = 5$ .

Check:

(g \circ f)(5) = g(f(5)) = g(2) = 10 = c

So the proof produces $b = 2$ and $x = 5$ .

Problem 2 · The First Line of the Proof

Given: $f\colon A \to B$ and $g\colon B \to C$ are surjections. Which sentence correctly begins a proof that $g \circ f\colon A \to C$ is a surjection?

✅ Correct! The goal

\forall c \in C,\ \exists x \in A,\ (g \circ f)(x) = c

starts with

\forall c

, so the proof chooses an arbitrary

c

and then has an existence statement to prove.

❌ That runs forward. Starting from an input only shows that its output lies in

C

, which is true of every function into

C

; it never shows that a given

c

is hit.

❌ Wrong quantifier on $x$ . This would prove

\forall x\, \forall c,\ (g \circ f)(x) = c

, which says every input lands on every element of

C

. The input must be found, not chosen:

\exists x

❌ That assumes the conclusion. The hypotheses are that

f

and

g

are surjections; that

g \circ f

is a surjection is what has to be proved.

❌ Not quite. Look at the first quantifier of the goal.

Show solution

Since $g \circ f$ has codomain $C$ , " $g \circ f$ is a surjection" means

\forall c \in C,\ \exists x \in A \text{ such that } (g \circ f)(x) = c.

The statement begins with a universal quantifier, so the choose-an-element method applies: let $c \in C$ be arbitrary, and the remaining goal is the existence statement $\exists x \in A$ with $(g \circ f)(x) = c$ . That is option b.

Option a starts at an input and goes forward, which only shows some elements of $C$ are outputs. Option c fixes $x$ in advance, so it would need every input to land on every $c$ . Option d assumes the very statement being proved.

Problem 3 · More Than One Choice

Given: $f\colon \{1, 2, 3, 4\} \to \{p, q, r\}$ with $f(1) = p$ , $f(2) = q$ , $f(3) = q$ , $f(4) = r$ , and $g\colon \{p, q, r\} \to \{u, v\}$ with $g(p) = u$ , $g(q) = u$ , $g(r) = v$ . Both are surjections. The proof pulls $c = u$ back to some $b$ with $g(b) = u$ , then to some $x$ with $f(x) = b$ .

How many choices are there for $b$ ?

How many different inputs $x$ can the proof end with?

✅ Correct!

b \in \{p, q\}

and

x \in \{1, 2, 3\}

; each of

1

2

3

satisfies

(g \circ f)(x) = u

, and the proof may use any of them.

❌ "There exists" is not "there is exactly one". Both

g(p) = u

and

g(q) = u

, so

g

being onto guarantees at least one

b

, and here there are two.

❌ Check $b$ . List the elements of

\{p, q, r\}

that

g

sends to

u

❌ That counts the choices of $b$ . The choice

b = q

has two inputs, since

f(2) = q

and

f(3) = q

; add the one input

1

for

b = p

❌ Check $x$ . For each possible

b

, list the inputs with

f(x) = b

Show solution

Through $g$ : the elements with $g(b) = u$ are $p$ and $q$ , so there are $2$ choices for $b$ .

Through $f$ : for $b = p$ the only input is $1$ ; for $b = q$ the inputs are $2$ and $3$ . So the proof can end with $x = 1$ , $2$ or $3$ : three inputs.

Check: $(g \circ f)(1) = g(p) = u$ , $(g \circ f)(2) = g(q) = u$ , $(g \circ f)(3) = g(q) = u$ . Input $4$ is not possible, since $(g \circ f)(4) = g(r) = v$ .

The proof needs no formula for $b$ or $x$ : any element that the two surjections guarantee will do.

Problem 4 · Three Surjections in a Row

Given: surjections $f\colon A \to B$ , $g\colon B \to C$ and $h\colon C \to D$ . A proof that $h \circ g \circ f\colon A \to D$ is onto lets $d \in D$ be arbitrary. Which sequence of steps should follow?

✅ Correct! Pull

d

back through the outermost function

h

, then

g

, then the innermost

f

; then

(h \circ g \circ f)(x) = h(g(b)) = h(c) = d

❌ The elements appear before they exist. The first step uses

b

, but

b

is only produced by the second step, and

c

only by the third. The search has to start at

d

and work inward.

❌ $g$ does not reach $D$ . The codomain of

g

C

, so

g(b) = d

makes no sense; the surjection

h

is needed first.

❌ That runs forward. It only shows the outputs lie in

D

, never that the given

d

is one of them.

❌ Not quite. Start from

d

and undo the outermost function first.

Show solution

Let $d \in D$ be arbitrary.

$h$ is onto $D$ , so there exists $c \in C$ with $h(c) = d$ .
$g$ is onto $C$ , so there exists $b \in B$ with $g(b) = c$ .
$f$ is onto $B$ , so there exists $x \in A$ with $f(x) = b$ .

(h \circ g \circ f)(x) = h(g(f(x))) = h(g(b)) = h(c) = d

Since $d$ was arbitrary, $h \circ g \circ f$ is a surjection. Equivalently, apply the theorem twice: $g \circ f$ is onto $C$ , and then $h \circ (g \circ f)$ is onto $D$ .

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