# Equivalence of Definitions of Composition of Mappings

## Theorem

The following definitions of the concept of Composition of Mappings are equivalent:

Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings such that the domain of $f_2$ is the same set as the codomain of $f_1$.

### Definition 1

The composite mapping $f_2 \circ f_1$ is defined as:

$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$

### Definition 2

The composite of $f_1$ and $f_2$ is defined and denoted as:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$

### Definition 3

The composite of $f_1$ and $f_2$ is defined and denoted as:

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

## Proof

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

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

### $(1)$ implies $(2)$

Let $f_2 \circ f_1$ be a composite mapping by definition 1.

Then by definition:

$\forall x \in S_1: \map {\paren {f_2 \circ f_1} } x := \map {f_2} {\map {f_1} x}$

We have that $f_1$ is a mapping, and so:

$\forall x \in S_1: \exists \map {f_1} x \in S_2$

We have that $f_2$ is a mapping, and so:

$\forall \map {f_1} x \in S_2: \exists \map {f_2} {\map {f_1} x} \in S_3$

Then:

 $\ds z$ $=$ $\ds \map {f_2} {\map {f_1} x}$ $\ds \leadsto \ \$ $\ds \tuple {\map {f_1} x, z}$ $\in$ $\ds f_2$ Definition of Mapping as a Relation

and so:

$f_2 \circ f_1 = \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$

Thus $f_2 \circ f_1$ is a composite mapping by definition 2.

$\Box$

### $(2)$ implies $(1)$

Let $f_2 \circ f_1$ be a composite mapping by definition 2.

Then by definition:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$

We have that $f_1$ is a mapping, and so:

$\forall x \in S_1: \exists \map {f_1} x \in S_2$

and so by definition of a Definition of Mapping as a Relation:

 $\ds \forall x \in S_1: \,$ $\ds \tuple {\map {f_1} x, z}$ $\in$ $\ds f_2$ $\ds \leadsto \ \$ $\ds z$ $=$ $\ds \map {f_2} {\map {f_1} x}$

Thus $f_2 \circ f_1$ is a composite mapping by definition 1.

$\Box$

### $(2)$ implies $(3)$

Let $f_2 \circ f_1$ be a composite mapping by definition 2.

Then by definition:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$

Because $f_1$ is a mapping, it follows that:

$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$

Similarly:

$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$

Hence:

$\tuple {\map {f_1} x, z} \in f_2 \implies \exists y \in S_2: \map {f_1} x = y \land \map {f_2} y = z$

and so:

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

Thus $f_2 \circ f_1$ is a composite mapping by definition 3.

$\Box$

### $(3)$ implies $(2)$

Let $f_2 \circ f_1$ be a composite mapping by definition 3.

Then by definition:

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

We have that:

$\forall x \in S_1: \exists y \in S_2: \map {f_1} x = y$

and that:

$\forall y \in S_2: \exists z \in S_3: \map {f_2} y = z$

Hence:

$\forall x \in S_1: \exists z \in S_3: \map {f_1} x = y \land \map {f_2} y = z$

Thus:

$\forall x \in S_1: \exists z \in S_3: \tuple {\map {f_1} x, y} \in f_2$

and so:

$f_2 \circ f_1 := \set {\tuple {x, z} \in S_1 \times S_3: \tuple {\map {f_1} x, z} \in f_2}$

Thus $f_2 \circ f_1$ is a composite mapping by definition 2.

$\blacksquare$