Mathematical-Reasoning · Unit 24 · Video 4 · Interactive Practice

Proof: f Inverse Is a Function Exactly When f Is a Bijection

IKey Formulas

Statement	Meaning
$f(x) = b \iff (x, b) \in f \iff (b, x) \in f^{-1}$	The chain used at every step of the proof
$f^{-1}$ is a function from $B$ to $A$ $\iff$ $f$ is a bijection	The theorem, for nonempty sets $A$ , $B$ and $f : A \to B$
$f$ surjection $\iff$ condition (1) for $f^{-1}$	Existence: every $b \in B$ is the first coordinate of at least one pair of $f^{-1}$
$f$ injection $\iff$ condition (2) for $f^{-1}$	Uniqueness: $(b, x_1), (b, x_2) \in f^{-1}$ implies $x_1 = x_2$

Key Insight: Existence matches existence and uniqueness matches uniqueness. The forward direction uses surjectivity for condition (1) and injectivity for condition (2); the converse runs the same correspondence backward.

IIWhich Property Fixes Which Condition?

Each failure of $f$ shows up as one specific failed condition for $f^{-1}$ , at one specific element of $B$ .

Size of B

💡 Challenge: with $3$ or $5$ elements in $B$ , no choice of arrows makes both checks pass. Why must a bijection between finite sets have equal sizes?

IIIThe Proof, Step by Step

Every step of the proof is one link of the chain, read in one direction or the other.

Step 1

The chain.

f(x) = b \iff (x, b) \in f \iff (b, x) \in f^{-1}

IVQuiz Questions

Problem 1 · Which Hypothesis Does the Work?

Given: the forward direction, where $f$ is a bijection. Which property of $f$ proves that every element of $B$ is the first coordinate of at least one pair of $f^{-1}$ ?

✅ Correct! For

b \in B

, surjectivity gives

x

with

f(x) = b

, hence

(b, x) \in f^{-1}

: existence from existence.

❌ Injectivity gives uniqueness. It shows that

b

is the first coordinate of at most one pair, which is condition (2).

❌ Being a function concerns $A$ . It makes every element of

A

an input of

f

; it says nothing about which elements of

B

are outputs.

❌ That is what we are proving. In the forward direction, "

f^{-1}

is a function" is the conclusion, so it cannot be used.

Show solution

Condition (1) for $f^{-1}$ asks: for every $b \in B$ , is there an $x \in A$ with $(b, x) \in f^{-1}$ , that is, with $f(x) = b$ ? That is word for word the definition of $f$ being a surjection.

Problem 2 · The Squaring Function

Given: $f : \mathbb{R} \to \mathbb{R}$ , $f(x) = x^2$ . Which conditions does $f^{-1}$ fail as a possible function from $\mathbb{R}$ to $\mathbb{R}$ ?

✅ Correct!

f

is neither an injection (

f(2) = f(-2)

) nor a surjection (no

x

has

x^2 = -1

❌ Only half the story. That witness is right, but

f

fails both properties, so

f^{-1}

fails both conditions.

❌ Not quite.

\sqrt{\ }

is the inverse of squaring only after restricting to

x \ge 0

and outputs

\ge 0

Show solution

$f^{-1} = \{ (y, x) \in \mathbb{R} \times \mathbb{R} \mid y = x^2 \}$ .

Condition (2) fails: $f(2) = f(-2) = 4$ , so $(4, 2)$ and $(4, -2)$ are both in $f^{-1}$ with $2 \ne -2$ . ( $f$ is not an injection.)
Condition (1) fails: $x^2 \ge 0$ for every real $x$ , so no pair of $f^{-1}$ has first coordinate $-1$ . ( $f$ is not a surjection.)

Problem 3 · An Injection on the Integers

Given: $f : \mathbb{Z} \to \mathbb{Z}$ , $f(n) = 2n$ . Which statement about $f^{-1}$ is correct?

✅ Correct!

f

is an injection, so no first coordinate repeats; but the odd integers are not outputs of

f

, so they start no pair.

❌ Swapped.

2m = 2n

forces

m = n

, so condition (2) holds; the trouble is the odd integers.

❌ Check the codomain.

\tfrac{3}{2}

is not an integer, so

n \mapsto \tfrac{n}{2}

is not a function from

\mathbb{Z}

\mathbb{Z}

❌ Not quite. Test injectivity and surjectivity of

f

separately.

Show solution

$f^{-1} = \{ (2n, n) \mid n \in \mathbb{Z} \}$ . Condition (2): if $(2m, m)$ and $(2n, n)$ have the same first coordinate, then $2m = 2n$ and $m = n$ , so it holds. Condition (1): $3 = 2n$ has no integer solution, so $3$ is the first coordinate of no pair, and it fails. This matches the theorem: $f$ is an injection but not a surjection.

Problem 4 · Justifying a Step

Given: this step of the converse, where we assume $f^{-1}$ is a function from $B$ to $A$ : "Let $f(x_1) = f(x_2) = b$ . Then $(b, x_1)$ and $(b, x_2)$ are in $f^{-1}$ . Therefore $x_1 = x_2$ ." Which reason justifies the word "therefore"?

✅ Correct! Condition (2) for the function

f^{-1}

: two pairs with first coordinate

b

have the same second coordinate.

❌ Circular. This step is proving that

f

is an injection, so it cannot assume it.

❌ Wrong condition. "Every element is an input" is condition (1); it gives existence, which is used for surjectivity, not uniqueness.

❌ Not quite. The definition only produces the two pairs; something must force their second coordinates to agree.

Show solution

The definition of $f^{-1}$ turns $f(x_1) = b$ and $f(x_2) = b$ into $(b, x_1), (b, x_2) \in f^{-1}$ . Because $f^{-1}$ is assumed to be a function, its input $b$ has only one output (condition (2)), so $x_1 = x_2$ . That proves $f$ is an injection.

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