# Domain of Composite Mapping

## Theorem

Let $S_1, S_2, S_3$ be sets.

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.

Let $f_2 \circ f_1: S_1 \to S_3$ be the composite mapping of $f_1$ and $f_2$.

Then:

$\Dom {f_1} = \Dom {f_2 \circ f_1}$

where $\Dom {f_1}$ denotes the domain of $f_1$.

## Proof

By definition of composition of mappings:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \exists y \in S_2: \tuple {x, y} \in f_1 \land \tuple {y, z} \in f_2}$

Let $x \in \Dom {f_2 \circ f_1}$.

Then:

$\exists z \in S_3: \tuple {x, z} \in S_1 \times S_3$

and:

$\exists y \in S_2: \tuple {x, y} \in f_1$

That is:

$x \in \Dom {f_1}$

Thus by definition of subset:

$\Dom {f_2 \circ f_1} \subseteq \Dom {f_1}$

Now suppose $x \in \Dom {f_1}$.

By the definition of mapping:

$\exists y \in S_2: \tuple {x, y} \in f_1$

As $f_2$ is likewise a mapping:

$\exists z \in S_3: \tuple {y, z} \in f_2$

and so by definition of composition of mappings:

$\tuple {x, z} \in S_1 \times S_3$

and so:

$x \in \Dom {f_2 \circ f_1}$

Thus by definition of subset:

$\Dom {f_1} \subseteq \Dom {f_2 \circ f_1}$

By definition of set equality:

$\Dom {f_1} = \Dom {f_2 \circ f_1}$

$\blacksquare$