# Image of Intersection under Injection/Family of Sets

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## Theorem

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:

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

if and only if $f$ is an injection.

## Proof

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: Family of Sets applies:

- $\displaystyle \mathcal R \sqbrk {\bigcap_{i \mathop \in I} S_i} = \bigcap_{i \mathop \in I} \mathcal R \sqbrk {S_i}$

if and only if $\mathcal R$ 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:

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

if and only if $f$ is an injection.

$\blacksquare$

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 6$. Indexed families; partitions; equivalence relations: Exercise $1$ - 1975: Bert Mendelson:
*Introduction to Topology*(3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $3 \ \text{(c})$