Condition for Composite Mapping on Left

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

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

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

Then $$\mathcal{R}$$ may be a mapping iff:
 * $$\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$$

That is:
 * $$\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$$

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

Sufficient Condition
Suppose $$\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$$.

Consider the subset $$G \subseteq \operatorname{Im} \left({f}\right) \times C$$ defined by:
 * $$G = \left\{{\left({y, z}\right): \exists x \in A: y = f \left({x}\right), z = g \left({x}\right)}\right\}$$

Clearly $$G \ne \varnothing$$ because for any $$x \in A$$ we have $$\left({f \left({x}\right), g \left({x}\right)}\right) \in G$$.

What we need to show is that $$G$$ is the graph of a mapping $$t: \operatorname{Im} \left({f}\right) \to C$$.

To do that, we need to show that for every $$y \in \operatorname{Im} \left({f}\right)$$, there is a unique $$z \in C$$ such that $$\left({y, z}\right) \in G$$.

It is clear that there is at least one such $$z$$: choose any $$x \in A$$ such that $$y = f \left({x}\right)$$ and set $$z = g \left({x}\right)$$.

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

Suppose we have $$\left({y, z}\right) \in G$$ and $$\left({y, z'}\right) \in G$$.

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

We have taken as a hypothesis that:
 * $$\forall x, y \in A: f \left({x}\right) = f \left({y}\right) \implies g \left({x}\right) = g \left({y}\right)$$

So $$g \left({x}\right) = g \left({x'}\right)$$ and so $$z = z'$$.

So $$G$$ is the graph of a mapping which we can denote:
 * $$t: \operatorname{Im} \left({f}\right) \to C$$

Also, since:
 * $$\forall x \in A: \left({f \left({x}\right), g \left({x}\right)}\right) \in G$$

it follows that:
 * $$\forall x \in A: f \left({x}\right) = t \left({f \left({x}\right)}\right)$$

We can now construct a mapping $$h$$ as follows:
 * $$h \left({x}\right) = \begin{cases}

t \left({x}\right) & : x \in \operatorname{Im} \left({f}\right) \\ \text {any } \alpha \in C & : x \in B - \operatorname{Im} \left({f}\right) \end{cases}$$

So:
 * $$\forall x \in A: \left({h \circ f}\right) \left({x}\right) = h \left({f \left({x}\right)}\right) = t \left({f \left({x}\right)}\right) = g \left({x}\right)$$

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 $$f \left({x}\right) = f \left({y}\right)$$. Then:

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.