Image of Intersection under Injection/Proof 2

Proof
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:
 * $f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$

$f$ 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.

$f$ 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.

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:

So if $f$ is an injection, it follows that:
 * $f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$

Putting the results together, we see that:
 * $f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ $f$ is an injection.