Image of Intersection under Injection/Proof 2

Proof
From Image of Intersection under Mapping:
 * $f \left[{A \cap B}\right] \subseteq f \left[{A}\right] \cap f \left[{B}\right]$

which holds for all mappings.

It remains to be shown that:
 * $f \left[{A}\right] \cap f \left[{B}\right] \subseteq f \left[{A \cap B}\right]$

$f$ is an injection.

Sufficient Condition
Suppose that:
 * $\forall A, B \subseteq S: f \left[{A \cap B}\right] = f \left[{A}\right] \cap f \left[{B}\right]$

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

So, assume $S$ is not singleton.

Suppose $f$ is specifically not an injection.

Then:
 * $\exists x, y \in S: \exists z \in T: \left({x, z}\right) \in T, \left({y, z}\right) \in T, x \ne y$

and of course $\left\{{x}\right\} \subseteq S, \left\{{y}\right\} \subseteq S$.

So:
 * $z \in f \left[{\left\{{x}\right\}}\right]$
 * $z \in f \left[{\left\{{y}\right\}}\right]$

and so by definition of intersection:
 * $z \in f \left[{\left\{{x}\right\}}\right] \cap f \left[{\left\{{y}\right\}}\right]$

But:
 * $\left\{{x}\right\} \cap \left\{{y}\right\} = \varnothing$

Thus from Image of Empty Set is Empty Set:
 * $\mathcal R \left[{\left\{{x}\right\} \cap \left\{{y}\right\}}\right] = \varnothing$

and so:
 * $\mathcal R \left[{\left\{{x}\right\} \cap \left\{{y}\right\}}\right] \ne f \left[{\left\{{x}\right\}}\right] \cap \mathcal R \left[{\left\{{y}\right\}}\right]$

Necessary Condition
Let $f$ be an injection.

It is necessary to show:


 * $f \left[{S_1}\right] \cap f \left[{S_2}\right] \subseteq f \left[{S_1 \cap S_2}\right]$

Let $t \in f \left[{S_1}\right] \cap f \left[{S_2}\right]$.

Then:

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

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