Mathematical-Reasoning · Unit 25 · Video 5 · Interactive Practice

Preimages Preserve Unions and Intersections Exactly

IKey Formulas

Result	Kind	Reason in one line
$f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$	Equality	Membership of $x$ depends on the single value $f(x)$
$f^{-1}(C \cup D) = f^{-1}(C) \cup f^{-1}(D)$	Equality	The same chain, with "or" in place of "and"
$A \subseteq f^{-1}(f(A))$	Inclusion	$x \in A$ gives $f(x) \in f(A)$
$f(f^{-1}(C)) \subseteq C$	Inclusion	An element of the image comes from some $x$ with $f(x) \in C$

Key Insight: $f(f^{-1}(C))$ is the part of $C$ that $f$ actually reaches, and $f^{-1}(f(A))$ adds to $A$ every input that shares an output with an element of $A$ . Either round trip can change the set.

IIThe If-and-Only-If Chain for One Input

Every line of the chain has the same truth value, because each step only unpacks a definition.

Operation Input x 3

x \in f^{-1}(C \cap D)

start

\iff f(x) \in C \cap D

definition of preimage

\iff f(x) \in C

and

f(x) \in D

definition of intersection

\iff x \in f^{-1}(C)

and

x \in f^{-1}(D)

definition of preimage, twice

\iff x \in f^{-1}(C) \cap f^{-1}(D)

definition of intersection

x \in f^{-1}(C \cup D)

start

\iff f(x) \in C \cup D

definition of preimage

\iff f(x) \in C

f(x) \in D

definition of union

\iff x \in f^{-1}(C)

x \in f^{-1}(D)

definition of preimage, twice

\iff x \in f^{-1}(C) \cup f^{-1}(D)

definition of union

Here $f(x) = (x^2 + 2) \bmod 8$ on $R_8$ , with $C = \{1, 2, 3\}$ and $D = \{3, 4, 5\}$ , as in the video's modulo $8$ example.

IIIWhere the Round Trips Leak

$f^{-1}(f(A))$ grows past $A$ when an outside input shares an output; $f(f^{-1}(C))$ shrinks below $C$ when $C$ holds a non-output.

Clicking

IVQuiz Questions

Problem 1 · Justify the Middle Step

Given: the chain proving $f^{-1}(C \cap D) = f^{-1}(C) \cap f^{-1}(D)$ contains the step " $f(x) \in C$ and $f(x) \in D$ $\iff$ $x \in f^{-1}(C)$ and $x \in f^{-1}(D)$ ." What justifies this step?

✅ Correct!

x \in f^{-1}(C)

means exactly

f(x) \in C

, and likewise for

D

; no property of

f

is needed.

❌ Not quite. Neither side of this step mentions an intersection; the definition of intersection is used in the steps before and after it.

❌ Not quite. The theorem holds for every function. The step only rewrites

f(x) \in C

x \in f^{-1}(C)

❌ Not quite. No image appears in this step:

f(x)

is a single output, and the sets involved are preimages.

Show solution

By definition, $f^{-1}(C) = \{ x \in S \mid f(x) \in C \}$ , so for $x \in S$ , $x \in f^{-1}(C) \iff f(x) \in C$ ; the same holds with $D$ . Replacing each half of the "and" statement by its equivalent gives the step. The definition reads in both directions, which is why the whole proof can be a chain of $\iff$ .

Problem 2 · Image, Then Preimage

Given: $g: \mathbb{R} \to \mathbb{R}$ , $g(x) = x^2$ , and $A = [0, 2]$ . Find $g^{-1}(g(A))$ .

✅ Correct!

g(A) = [0, 4]

, and

x^2 \in [0, 4]

exactly when

-2 \le x \le 2

. Each negative input shares its square with a positive one in

A

❌ Not quite. Image and preimage need not cancel:

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

with

1 \in A

, so

-1 \in g^{-1}(g(A))

❌ Close, but… that is

g(A)

, the halfway point. Now take its preimage.

❌ Not quite.

g(3) = 9 \notin [0, 4]

, so

3

is not in the preimage; the bound on

x

\sqrt{4} = 2

Show solution

Image: on $[0, 2]$ the squares rise from $0$ to $4$ , so $g(A) = [0, 4]$ .

Preimage: $x \in g^{-1}([0, 4])$ exactly when $0 \le x^2 \le 4$ , that is, $-2 \le x \le 2$ .

g^{-1}(g(A)) = [-2, 2] \supsetneq [0, 2] = A.

Problem 3 · Preimage, Then Image

Given: $h: \mathbb{R} \to \mathbb{R}$ , $h(x) = |x|$ , and $C = [-3, 2]$ . Find $h(h^{-1}(C))$ .

✅ Correct!

h^{-1}(C) = [-2, 2]

, and

|x|

[-2, 2]

takes exactly the values

[0, 2]

. The negative part of

C

is never an output.

❌ Not quite.

f(f^{-1}(C))

is only the part of

C

that the function reaches, and

|x|

never takes a negative value.

❌ Close, but… that is

h^{-1}(C)

. Apply

h

to it.

❌ Not quite.

h(1) = 1 \in C

, so

1 \in h^{-1}(C)

and

1 \in h(h^{-1}(C))

Show solution

Preimage: $|x| \in [-3, 2]$ means $-3 \le |x| \le 2$ ; the left inequality always holds, so the condition is $|x| \le 2$ , and $h^{-1}(C) = [-2, 2]$ .

Image: as $x$ runs over $[-2, 2]$ , $|x|$ takes every value from $0$ to $2$ , so

h(h^{-1}(C)) = [0, 2] \subsetneq [-3, 2] = C.

Problem 4 · Reading a Gap

Given: $f: S \to T$ and $C \subseteq T$ such that some element $c \in C$ is not $f(x)$ for any $x \in S$ . Which statement must be true?

✅ Correct! Every element of

f(f^{-1}(C))

is an output of

f

, and

c

is not, so

c \in C

but

c \notin f(f^{-1}(C))

❌ Not quite. Other elements of

C

may be outputs: for

f(x) = (x^2 + 2) \bmod 8

R_8

C = \{1, 2, 3\}

contains the non-output

1

, yet

f^{-1}(C) = \{0, 1, 3, 4, 5, 7\}

❌ Not quite. The element

c

cannot come back: anything in

f(f^{-1}(C))

has the form

f(x)

❌ Not quite. This concerns the other round trip, and it fails for

A = S

f^{-1}(f(S)) = S

for every function.

Show solution

If $y \in f(f^{-1}(C))$ , then $y = f(x)$ for some $x \in f^{-1}(C)$ , so $y$ is an output of $f$ . The element $c$ is not an output, so $c \notin f(f^{-1}(C))$ while $c \in C$ . Hence $f(f^{-1}(C)) \ne C$ , and with $f(f^{-1}(C)) \subseteq C$ the inclusion is proper.

(a) fails whenever $C$ also contains an output; (d) fails for $A = S$ , since every input lies in $f^{-1}(f(S))$ and that set is a subset of $S$ .

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