# Condition for Composite Mapping on Right

## Theorem

Let $A, B, C$ be sets.

Let $f: B \to A$ and $g: C \to A$ be mappings.

Let $\RR: C \to B$ be a relation such that $g = f \circ \RR$ is the composite of $\RR$ and $f$.

Then $\RR$ may be a mapping if and only if:

- $\Img g \subseteq \Img f$

That is:

- $\Img g \subseteq \Img f$

- $\exists h: C \to B$ such that $h$ is a mapping and $f \circ h = g$

## Proof

### Sufficient Condition

Suppose $\Img g \subseteq \Img f$.

That is:

- $\forall x \in C: \map g x \in \Img f$

and so:

- $\forall x \in C: \exists y \in B: \map g x = \map f y$

Take any $x \in C$.

Consider the set $Y_x = \set {y \in B: \map g x = \map f y}$.

We know from above that $Y_x \ne \O$.

So, using the Axiom of Choice, for each $x$ we may select some $y_x \in Y_x$.

Then we may define the mapping $h: C \to B$ such that:

- $\forall x \in C: \map h x = y_x$

We then see that:

\(\ds \forall x \in C: \, \) | \(\ds \map {\paren {f \circ h} } x\) | \(=\) | \(\ds \map f {\map h x}\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \map f {y_x}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \map g x\) |

Thus we have constructed a mapping $h$ such that $f \circ h = g$, as required.

### Necessary Condition

Suppose there exists some mapping $h: C \to B$ such that $f \circ h = g$.

Let $y \in \Img g$.

Then we have:

\(\ds y\) | \(\in\) | \(\ds \Img g\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds \exists x \in C: \, \) | \(\ds \map g x\) | \(=\) | \(\ds y\) | Definition of Image of Mapping | |||||||||

\(\ds \leadsto \ \ \) | \(\ds \map {\paren {f \circ h} } x\) | \(=\) | \(\ds y\) | Definition of $g$ | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map f {\map h x}\) | \(=\) | \(\ds y\) | Definition of Composition of Mappings | ||||||||||

\(\ds \leadsto \ \ \) | \(\ds y\) | \(\in\) | \(\ds \Img f\) | Definition of Image of Mapping |

Hence by definition of subset, $\Img g \subseteq \Img f$.

$\blacksquare$

## Comment

Hence we have a necessary and sufficient condition for determining whether the composition of mappings actually exists as a mapping.

Note that this is different from being given two mappings and creating their composition.

## Axiom of Choice

This theorem depends on the Axiom of Choice.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

## Sources

- 1975: T.S. Blyth:
*Set Theory and Abstract Algebra*... (previous) ... (next): $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.5$