Theorem: Image of Intersection under Injection

Let $S$ and $T$ be sets.
Let $f: S \to T$ be a mapping.
Then:
:$\forall A, B \subseteq S: f \sqbrk {A \cap B} = f \sqbrk A \cap f \sqbrk B$
{{iff}} $f$ is an injection.
This can be expressed in the language and notation of direct image mappings as:
:$\forall A, B \subseteq S: \map {f^\to} {A \cap B} = \map {f^\to} A \cap \map {f^\to} B$

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:
:$f \sqbrk A \cap f \sqbrk B \subseteq f \sqbrk {A \cap B}$
{{iff}} $f$ 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.
{{AimForCont}} $f$ 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 {{Corollary|Image of Empty Set is Empty Set|1}}:
:$f \sqbrk {\set x \cap \set y} = \O$
and so:
:$f \sqbrk {\set x \cap \set y} \ne f \sqbrk {\set x} \cap f \sqbrk {\set y}$
But {{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.
{{qed|lemma}}
=== 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:
{{begin-eqn}}
{{eqn | l = t
      | o = \in
      | r = f \sqbrk {S_1} \cap f \sqbrk {S_2}
      | c =
}}
{{eqn | ll= \leadsto
      | l = t
      | o = \in
      | r = f \sqbrk {S_1}
      | c = {{Defof|Set Intersection}}
}}
{{eqn | lo= \land
      | l = t
      | o = \in
      | r = f \sqbrk {S_2}
      | c =
}}
{{eqn | ll= \leadsto
      | q = \exists s_1 \in S_1, s_2 \in S_2
      | l = t
      | r = \map f {s_1}
      | c = {{Defof|Image of Subset under Mapping}}
}}
{{eqn | lo= \land
      | l = t
      | r = \map f {s_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map f {s_1}
      | r = \map f {s_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = s_1
      | r = s_2
      | c = {{Defof|Injection}}
}}
{{eqn | ll= \leadsto
      | l = s_1 = s_2
      | o = \in
      | r = S_1
      | c = 
}}
{{eqn | lo= \land
      | l = s_1 = s_2
      | o = \in
      | r = S_2
      | c = 
}}
{{eqn | ll=\leadsto
      | l = s_1 = s_2
      | o = \in
      | r = S_1 \cap S_2
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | l = \map f {s_1} = \map f {s_2}
      | o = \in
      | r = f \sqbrk {S_1 \cap S_2}
      | c = Image of Element is Subset
}}
{{eqn | ll= \leadsto
      | l = f \sqbrk {S_1} \cap f \sqbrk {S_2}
      | o = \subseteq
      | r = f \sqbrk {S_1 \cap S_2}
      | c = {{Defof|Subset}}
}}
{{end-eqn}}
So if $f$ is an injection, it follows that:
:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$
{{qed|lemma}}
Putting the results together, we see that:
:$f \sqbrk {S_1 \cap S_2} = f \sqbrk {S_1} \cap f \sqbrk {S_2}$ {{iff}} $f$ is an injection.
{{qed}}