# Preimage of Union Mapping is Union of Preimages

## Theorem

Let $A$, $B$ and $Y$ be sets.

Let $f: A \to Y$ and $g: B \to Y$ be mappings that agree on $A \cap B$.

Let $f \cup g$ be the union of the mappings $f$ and $g$:

$\forall x \in A \cup B: \map {f \cup g} x = \begin {cases} \map f x & : x \in A \\ \map g x & : x \in B \end {cases}$

Let $S \subseteq Y$ be a subset of $Y$.

Then:

$\paren {f \cup g}^{-1} \sqbrk S = f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$

## Proof

$f \cup g$ is well-defined

Let $x \in \paren {f \cup g}^{-1} \sqbrk S$

By Definition of Inverse Image:

$\paren {f \cup g}^{-1} \sqbrk S \subseteq A \cup B$

Let $x \in A$.

We have:

$\map f x = \map {\paren {f \cup g} } x \in S$

By Definition of Inverse Image:

$x \in f^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$

Let $x \in B$.

We have:

$\map g x = \map {\paren {f \cup g} } x \in S$

By Definition of Inverse Image:

$x \in g^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$

In both cases:

$x \in f^{-1} \sqbrk U \cup g^{-1} \sqbrk S$

By Definition of Subset:

$\paren {f \cup g}^{-1} \sqbrk S \subseteq f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$

Let $x \in f^{-1} \sqbrk S \cup g^{-1} \sqbrk S$.

Without loss of generality consider the case of $x \in f^{-1} \sqbrk S \subseteq A$.

Then: $\map {\paren {f \cup g} } x = \map f x \in S$

Hence:

$x \in \paren {f \cup g}^{-1} \sqbrk S$

By Definition of Subset:

$f^{-1} \sqbrk S \cup g^{-1} \sqbrk S \subseteq \paren {f \cup g}^{-1} \sqbrk S$

By Definition of Set Equality:

$f^{-1} \sqbrk S \cup g^{-1} \sqbrk S = \paren {f \cup g}^{-1} \sqbrk S$

$\blacksquare$