Image of Intersection under Mapping

Theorem
The image of the intersection of subsets of a mapping is a subset of the intersection of their images.

That is:

Let $f: S \to T$ be a mapping. Let $S_1$ and $S_2$ be subsets of $S$.

Then:
 * $f \left({S_1 \cap S_2}\right) \subseteq f \left({S_1}\right) \cap f \left({S_2}\right)$

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

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

Let $\mathbb S \subseteq \mathcal P \left({S}\right)$.

Then:
 * $\displaystyle f \left({\bigcap \mathbb S}\right) \subseteq \bigcap_{X \in \mathbb S} f \left({X}\right)$

Proof
As $f$, being a mapping, is also a relation, we can apply Image of Intersection:


 * $\mathcal R \left({S_1 \cap S_2}\right) \subseteq \mathcal R \left({S_1}\right) \cap \mathcal R \left({S_2}\right)$

Note
Note that equality does not hold in general.

Let:
 * $S_1 = \left\{{x \in \Z: x \le 0}\right\}$
 * $S_2 = \left\{{x \in \Z: x \ge 0}\right\}$
 * $f: \Z \to \Z: \forall x \in \Z: f \left({x}\right) = x^2$

We have:
 * $f \left({S_1}\right) = \left\{{0, 1, 4, 9, 16, \ldots}\right\} = f \left({S_2}\right)$

Then:
 * $f \left({S_1}\right) \cap f \left({S_2}\right) = \left\{{0, 1, 4, 9, 16, \ldots}\right\}$

but:
 * $f \left({S_1 \cap S_2}\right) = f \left({\left\{{0}\right\}}\right) = \left\{{0}\right\}$

Note that from Injection Image of Intersections equality always holds iff $f$ is an injection.

Also see

 * One-to-Many Image of Intersections


 * Mapping Preimage of Intersection


 * Mapping Image of Union
 * Mapping Preimage of Union