Subset of Preimage under Relation is Preimage of Subset/Corollary

Corollary to Subset of Preimage under Relation is Preimage of Subset

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

Let $X \subseteq S, Y \subseteq T$.

Then:

$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$

This can be expressed in the language and notation of direct image mappings and inverse image mappings as:

$\forall X \in \powerset S, Y \in \powerset T: X \subseteq \map {f^\gets} Y \iff \map {f^\to} X \subseteq Y$

Proof

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

As a mapping is also a relation, it follows that $f$ is a relation and so:

$X \subseteq f^{-1} \sqbrk Y \iff f \sqbrk X \subseteq Y$

holds on the strength of Subset of Preimage under Relation is Preimage of Subset.

$\blacksquare$

Sources

• 1975: T.S. Blyth: Set Theory and Abstract Algebra ... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.1 \ \text{(v)}$