Retract of Injective Space is Injective

Theorem
Let $T = \left({S, \tau}\right)$ be an injective topological space.

Let $R = \left({Z, \tau'}\right)$ be a retract of $T$.

Then $R$ is injective.

Proof
By definition of retract:
 * there exists a continuous retraction $r: S \to Z$ of $T$.

Let $\mathcal Y = \left({Y, \sigma}\right)$ be a topological space.

Let $f: Y \to Z$ be a continuous mapping.

Let $\mathcal X = \left({X, \sigma'}\right)$ such that
 * $\mathcal Y$ is topological subspace of $\mathcal X$.

By Inclusion Mapping is Continuous:
 * $i_Z$ is continuous $\left({R \to T}\right)$

where $i_Z$ denotes the inclusion mapping from $Z$ in $S$.

By Composite of Continuous Mappings is Continuous:
 * $i_Z \circ f: Y \to S$ is continuous.

By definition of injective space:
 * there exists a continuous mapping $g:X \to S: g \restriction Y = i_Z \circ f$

Define $h := r \circ g$

By Composite of Continuous Mappings is Continuous:
 * $h$ is continuous.

We will prove that
 * $h \restriction Y = f$

By definition of topological subspace:
 * $Y \subseteq X$ and $Z \subseteq S$

Thus by definitions of composition of mappings and restriction of mapping:
 * $h \restriction Y: Y \to Z$ and $f: Y \to Z$

Let $y \in Y$.

By Restriction of Composition is Composition of Restriction:
 * $h \restriction Y = r \circ \left({g \restriction Y}\right)$

By definition of mapping:
 * $f\left({y}\right) \in Z$

Thus