Factoring Mapping into Surjection and Inclusion

Theorem
Every mapping $$f:S \to T$$ can be uniquely factored into a surjection $$g$$ followed by the inclusion mapping $$i$$.

That is, $$f = i \circ g$$ where:


 * $$g: S \to \operatorname{Im} \left({f}\right) : g \left({x}\right) = f \left({x}\right)$$
 * $$i: \operatorname{Im} \left({f}\right) \to T : i \left({x}\right) = x$$

Proof

 * From Surjection by Restriction of Range, any $$f: S \to \operatorname{Im} \left({f}\right)$$ is a surjection.

The mapping $$g: S \to \operatorname{Im} \left({f}\right)$$ where $$g \left({x}\right) = f \left({x}\right)$$ is therefore also clearly a surjection.

The mapping $$g: S \to \operatorname{Im} \left({f}\right) : g \left({x}\right) = f \left({x}\right)$$ is clearly unique, by Equality of Mappings.


 * From Inclusion Mapping is an Injection, $$i: \operatorname{Im} \left({f}\right) \to T$$ is an injection.

Likewise, the mapping $$i: \operatorname{Im} \left({f}\right) \to T : i \left({x}\right) = x$$ is also unique, by its own definition.