Condition for Composite Mapping on Left

Theorem
Let $A, B, C$ be sets.

Suppose that $C$ is non-empty.

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

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

Then $\RR$ may be a mapping :
 * $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$

That is:
 * $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$


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

Sufficient Condition
Suppose $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$.

Consider the subset $G \subseteq \Img f \times C$ defined by:
 * $G = \set {\tuple {y, z}: \exists x \in A: y = \map f x, z =\map g x}$

Clearly $G \ne \O$ because for any $x \in A$ we have $\tuple {\map f x, \map g x} \in G$.

What we need to show is that $G$ is the graph of a mapping $t: \Img f \to C$.

To do that, we need to show that for every $y \in \Img f$, there is a unique $z \in C$ such that $\tuple {y, z} \in G$.

It is clear that there is at least one such $z$: choose any $x \in A$ such that $y = \map f x$ and set $z = \map g x$.

Now we need to show that such a $z$ is unique.

Suppose we have $\tuple {y, z} \in G$ and $\tuple {y, z'} \in G$.

Then by the definition of $G$:
 * $\exists x, x' \in A: y = \map f x = \map f {x'}, z = \map g x, z' = \map g {x'}$

We have taken as a hypothesis that:
 * $\forall x, y \in A: \map f x = \map f y \implies \map g x = \map g y$

So $\map g x = \map g {x'}$ and so $z = z'$.

So $G$ is the graph of a mapping which we can denote:
 * $t: \Img f \to C$

Also, since:
 * $\forall x \in A: \tuple {\map f x, \map g x} \in G$

it follows that:
 * $\forall x \in A: \map f x = \map t {\map f x}$

Since $C$ is non-empty, it has an element $c$.

By Law of Excluded Middle, we can now construct a mapping $h$ as follows:
 * $\map h x = \begin{cases}

\map t x & : x \in \Img f \\ c        & : x \in B \setminus \Img f \end{cases}$

So:
 * $\forall x \in A: \map {\paren {h \circ f} } x = \map h {\map f x} = \map t {\map f x} = \map g x$

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

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

Let $\map f x = \map f y$.

Then we have:

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.