# Image of Intersection under Injection

## Contents

## Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Then:

- $\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$

if and only if $f$ is an injection.

### General Result

Let $S$ and $T$ be sets.

Let $f: S \to T$ be a mapping.

Let $\mathcal P \left({S}\right)$ be the power set of $S$.

Then:

- $\displaystyle \forall \mathbb S \subseteq \mathcal P \left({S}\right): f \left[{\bigcap \mathbb S}\right] = \bigcap_{X \mathop \in \mathbb S} f \left[{X}\right]$

if and only if $f$ is an injection.

### Family of Sets

Let $S$ and $T$ be sets.

Let $\family {S_i}_{i \mathop \in I}$ be a family of subsets of $S$.

Let $f: S \to T$ be a mapping.

Then:

- $\displaystyle f \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} f \sqbrk {S_i}$

if and only if $f$ is an injection.

## Proof 1

An injection is a type of one-to-one relation, and therefore also a one-to-many relation.

Therefore Image of Intersection under One-to-Many Relation applies:

- $\forall A, B \subseteq S: \mathcal R \left[{A \cap B}\right] = \mathcal R \left[{A}\right] \cap \mathcal R \left[{B}\right]$

if and only if $\mathcal R$ is a one-to-many relation.

We have that $f$ is a mapping and therefore a many-to-one relation.

So $f$ is a one-to-many relation if and only if $f$ is also an injection.

It follows that:

- $\forall A, B \subseteq S: f \left[{A \cap B}\right] = f \left[{A}\right] \cap f \left[{B}\right]$

if and only if $f$ is an injection.

$\blacksquare$

## Proof 2

From Image of Intersection under Mapping:

- $f \sqbrk {A \cap B} \subseteq f \sqbrk A \cap f \sqbrk B$

which holds for all mappings.

It remains to be shown that:

- $f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$

if and only if $f$ is an injection.

### Sufficient Condition

Suppose that:

- $\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$

If $S$ is singleton, then $f$ is a fortiori an injection from Mapping from Singleton is Injection.

So, assume $S$ is not singleton.

Aiming for a contradiction, suppose $f$ is specifically *not* an injection.

Then:

- $\exists x, y \in S: \exists z \in T: \tuple {x, z} \in T, \tuple {y, z} \in T, x \ne y$

and of course $\set x \subseteq S, \set y \subseteq S$.

So:

- $z \in f \sqbrk {\set x}$
- $z \in f \sqbrk {\set y}$

and so by definition of intersection:

- $z \in f \sqbrk {\set x} \cap f \sqbrk {\set y}$

But:

- $\set x \cap \set y = \O$

Thus from the corollary to Image of Empty Set is Empty Set:

- $f \sqbrk {\set x \cap \set y} = \O$

and so:

- $f \sqbrk {\set x \cap \set y} \ne f \sqbrk {\set x} \cap f \sqbrk {\set y}$

But by hypothesis:

- $\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$

From this contradiction it follows that $f$ is an injection.

$\Box$

### Necessary Condition

Let $f$ be an injection.

It is necessary to show:

- $f \sqbrk {S_1} \cap f \sqbrk {S_2} \subseteq f \sqbrk {S_1 \cap S_2}$

Let $t \in f \sqbrk {S_1} \cap f \sqbrk {S_2}$.

Then:

\(\displaystyle t\) | \(\in\) | \(\displaystyle f \sqbrk {S_1} \cap f \sqbrk {S_2}\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(\in\) | \(\displaystyle f \sqbrk {S_1}\) | Definition of Set Intersection | |||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle t\) | \(\in\) | \(\displaystyle f \sqbrk {S_2}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists s_1 \in S_1, s_2 \in S_2: \ \ \) | \(\displaystyle t\) | \(=\) | \(\displaystyle \map f {s_1}\) | Definition of Image of Subset under Mapping | ||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle t\) | \(=\) | \(\displaystyle \map f {s_2}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f {s_1}\) | \(=\) | \(\displaystyle \map f {s_2}\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1\) | \(=\) | \(\displaystyle s_2\) | Definition of Injection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1 = s_2\) | \(\in\) | \(\displaystyle S_1\) | ||||||||||

\(\, \displaystyle \land \, \) | \(\displaystyle s_1 = s_2\) | \(\in\) | \(\displaystyle S_2\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s_1 = s_2\) | \(\in\) | \(\displaystyle S_1 \cap S_2\) | Definition of Set Intersection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \map f {s_1} = \map f {s_2}\) | \(\in\) | \(\displaystyle f \sqbrk {S_1 \cap S_2}\) | Image of Element is Subset | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle f \sqbrk {S_1} \cap f \sqbrk {S_2}\) | \(\subseteq\) | \(\displaystyle f \sqbrk {S_1 \cap S_2}\) | Definition of Subset |

So if $f$ is an injection, it follows that:

- $f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$

$\Box$

Putting the results together, we see that:

- $f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ if and only if $f$ is an injection.

$\blacksquare$

## Sources

- 1960: Paul R. Halmos:
*Naive Set Theory*... (previous) ... (next): $\S 10$: Inverses and Composites: Exercise $\text{(ii)}$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $1$: Mappings: $\S 12 \alpha$ - 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $2$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $4$: Mappings: Exercise $7$ - 2000: James R. Munkres:
*Topology*(2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 2$: Functions: Exercise $2.2 \ \text{(g)}$