Retract of Injective Space is Injective

Theorem
Let $T = \struct {S, \tau}$ be an injective topological space.

Let $R = \struct {Z, \tau'}$ 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 $\YY = \struct {Y, \sigma}$ be a topological space.

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

Let $\XX = \struct {X, \sigma'}$ such that
 * $\YY$ is topological subspace of $\XX$.

By Inclusion Mapping is Continuous:
 * $i_Z$ is continuous $\paren {R \to T}$

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:
 * $h \restriction Y = r \circ \paren {g \restriction Y}$

By definition of mapping:
 * $\map f y \in Z$

Thus