Mathematical-Reasoning · Unit 24 · Video 2 · Interactive Practice

Reversing Every Arrow: The Function That Undoes a Bijection

IKey Formulas

Statement	Meaning
$f(a) = r,\ f(b) = p,\ f(c) = s,\ f(d) = q$	The video's bijection $f : \{a, b, c, d\} \to \{p, q, r, s\}$
$g(p) = b,\ g(q) = d,\ g(r) = a,\ g(s) = c$	The reversed arrows define $g : B \to A$ ; $g(y)$ is where the arrow of $f$ into $y$ started
$(g \circ f)(x) = x$ for all $x \in A$	$g \circ f$ is the identity function on $A$
$(f \circ g)(y) = y$ for all $y \in B$	$f \circ g$ is the identity function on $B$

Key Insight: Surjectivity gives every element of $B$ at least one arrow to reverse; injectivity makes it at most one. At least one and at most one is exactly one, so the reversed arrows form a function.

IIWhen Do the Reversed Arrows Form a Function?

Each element of $B$ sends out one reversed arrow for every arrow of $f$ it received.

IIIOut and Back Again

Following an arrow of $f$ and then the reversed arrow of $g$ returns every element to where it started.

Composite Input a

IVQuiz Questions

Problem 1 · Reading a Reversed Arrow

Given: the bijection $f : \{1, 2, 3\} \to \{u, v, w\}$ with $f(1) = w$ , $f(2) = u$ , $f(3) = v$ , and $g$ the function defined by the reversed arrows. What is $g(u)$ ?

✅ Correct! The arrow of

f

into

u

starts at

2

, so the reversed arrow from

u

ends at

2

❌ That reads the list by position.

u

is the output of

2

, not of

1

f(2) = u

❌ Wrong set.

g

goes from

\{u, v, w\}

back to

\{1, 2, 3\}

, so its outputs are numbers.

❌ Not quite. Find the element of

\{1, 2, 3\}

whose arrow lands on

u

Show solution

$g(u)$ is the start of the arrow of $f$ that ends at $u$ . Since $f(2) = u$ , the reversed arrow runs from $u$ to $2$ , so $g(u) = 2$ . Likewise $g(v) = 3$ and $g(w) = 1$ .

Problem 2 · Reversing a Non-Bijection

Given: $f : \{a, b, c\} \to \{p, q, r\}$ with $f(a) = p$ , $f(b) = p$ , $f(c) = q$ . Do the reversed arrows define a function from $\{p, q, r\}$ to $\{a, b, c\}$ ?

✅ Correct!

f

is not an injection (

a

and

b

p

) and not a surjection (

r

is missed), and each failure breaks the reversed arrows separately.

❌ That is why $f$ is a function. The reversed arrows start in

\{p, q, r\}

, so it is those elements that need exactly one arrow each.

❌ Only half the story. There are two separate failures here, not one.

Show solution

The reversed arrows are $p \to a$ , $p \to b$ and $q \to c$ .

$p$ sends out two arrows (to $a$ and to $b$ ), because two arrows of $f$ landed on $p$ : $f$ is not an injection.
$r$ sends out none, because no arrow of $f$ landed on $r$ : $f$ is not a surjection.

A function from $\{p, q, r\}$ needs exactly one arrow out of each element, so both $p$ and $r$ fail.

Problem 3 · The Other Composite

Given: the video's $f(a) = r$ , $f(b) = p$ , $f(c) = s$ , $f(d) = q$ and its reversed-arrow function $g$ . Which statement about $f \circ g$ is correct?

✅ Correct!

g

acts first and takes

B

A

; then

f

takes

A

back to

B

. And

f(g(q)) = f(d) = q

❌ Check the domain. In

f \circ g

the function

g

acts first, so the inputs are elements of

B

;

q

is not even in

A

❌ That stops halfway.

g(q) = d

is only the first step;

f

still has to act on

d

❌ Not quite. Apply

g

first, then

f

f(g(q)) = f(d)

Show solution

$f \circ g$ means " $g$ first, then $f$ ". Since $g : B \to A$ and $f : A \to B$ , the composite goes from $B$ to $B$ . At $q$ :

(f \circ g)(q) = f\big(g(q)\big) = f(d) = q.

The same happens at every $y \in B$ , so $f \circ g$ is the identity function on $B$ .

Problem 4 · What Reversal Requires

Given: finite sets $A$ , $B$ and a function $f : A \to B$ whose reversed arrows define a function from $B$ to $A$ . Which statement must be true?

✅ Correct! Every element of

B

received exactly one arrow, so

f

is onto and one-to-one, and the arrows pair the elements of

A

and

B

off one for one.

❌ A missed element breaks it. If some

y \in B

received no arrow, then

y

has no reversed arrow, so it has no output.

❌ A shared output breaks it. If two arrows land on

y

, then

y

sends out two reversed arrows.

❌ Not quite. Count arrows: each element of

B

must send out exactly one reversed arrow.

Show solution

Reversed arrows define a function from $B$ to $A$ exactly when each $y \in B$ sends out exactly one of them, that is, exactly one arrow of $f$ lands on each $y$ .

At least one arrow on each $y$ : $f$ is a surjection.
At most one arrow on each $y$ : $f$ is an injection.

So $f$ is a bijection. Its arrows match each element of $A$ with a different element of $B$ and use all of $B$ , so the two sets have the same number of elements.

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