# Image of Intersection under Injection

## 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 $\powerset S$ be the power set of $S$.

Then:

$\ds \forall \mathbb S \subseteq \powerset S: f \sqbrk {\bigcap \mathbb S} = \bigcap_{X \mathop \in \mathbb S} f \sqbrk X$

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:

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

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:

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

if and only if $\RR$ 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:

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

if and only if $f$ is an injection.

$\blacksquare$

## Proof 2

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

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 a 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:

 $\ds t$ $\in$ $\ds f \sqbrk {S_1} \cap f \sqbrk {S_2}$ $\ds \leadsto \ \$ $\ds t$ $\in$ $\ds f \sqbrk {S_1}$ Definition of Set Intersection $\, \ds \land \,$ $\ds t$ $\in$ $\ds f \sqbrk {S_2}$ $\ds \leadsto \ \$ $\ds \exists s_1 \in S_1, s_2 \in S_2: \,$ $\ds t$ $=$ $\ds \map f {s_1}$ Definition of Image of Subset under Mapping $\, \ds \land \,$ $\ds t$ $=$ $\ds \map f {s_2}$ $\ds \leadsto \ \$ $\ds \map f {s_1}$ $=$ $\ds \map f {s_2}$ $\ds \leadsto \ \$ $\ds s_1$ $=$ $\ds s_2$ Definition of Injection $\ds \leadsto \ \$ $\ds s_1 = s_2$ $\in$ $\ds S_1$ $\, \ds \land \,$ $\ds s_1 = s_2$ $\in$ $\ds S_2$ $\ds \leadsto \ \$ $\ds s_1 = s_2$ $\in$ $\ds S_1 \cap S_2$ Definition of Set Intersection $\ds \leadsto \ \$ $\ds \map f {s_1} = \map f {s_2}$ $\in$ $\ds f \sqbrk {S_1 \cap S_2}$ Image of Element is Subset $\ds \leadsto \ \$ $\ds f \sqbrk {S_1} \cap f \sqbrk {S_2}$ $\subseteq$ $\ds f \sqbrk {S_1 \cap S_2}$ 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.

$\blacksquare$