Retract of Injective Space is Injective

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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

\(\ds \map {\paren {h \restriction Y} } y\) \(=\) \(\ds \map r {\map {\paren {i_Z \circ f} } y}\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map r {\map {i_Z} {\map f y} }\) Definition of Composition of Mappings
\(\ds \) \(=\) \(\ds \map r {\map f y}\) Definition of Inclusion Mapping
\(\ds \) \(=\) \(\ds \map f y\) Definition of Retraction (Topology)

$\blacksquare$


Sources

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