Subspace of Product Space is Homeomorphic to Factor Space

Theorem

Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be a family of topological spaces where $I$ is an arbitrary index set.

Let $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.

Suppose that $X$ is non-empty.

Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $\struct {X_i, \tau_i}$.

Specifically, for any $z \in X$, let:

$Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$

and let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.

Then $\struct {Y_i, \upsilon_i}$ is homeomorphic to $\struct {X_i, \tau_i}$, where the homeomorphism is the restriction of the projection $\pr_i$ to $Y_i$.

Corollary

Let $T_1$ and $T_2$ be non-empty topological spaces.

Let $b \in T_2$.

Let $T_1 \times T_2$ be the product space of $T_1$ and $T_2$.

Let $T_2 \times T_1$ be the product space of $T_2$ and $T_1$.

Then:

$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$

$T_1$ is homeomorphic to the subspace $\set b \times T_1$ of $T_2 \times T_1$

Proof 1

Let $z \in X$.

Let $i \in i$.

Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.

Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.

For all $j \in I$ let:

$Z_j = \begin{cases} X_i & i = j \\ \set{z_j} & j \ne i \end{cases}$

Lemma 1

$Y_i = \prod_{j \mathop \in I} Z_j$

$\Box$

From Product Space of Subspaces is Subspace of Product Space , $\struct {Y_i, \upsilon_i}$ is a product space.

Consider the projection:

$p_i: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$

From Projection from Product of Family is Injection iff Other Factors are Singletons, $p_i$ is injective.

From Projection from Product of Family is Surjective, $p_i$ is surjective.

From Projection from Product Topology is Continuous:General Result, $p_i$ is continuous.

From Projection from Product Topology is Open:General Result, $p_i$ is open.

Thus, by definition, we have that $p_i$ is a homeomorphism.

Consider the projection:

$\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$

and the restriction:

$\pr_i {\restriction_{Y_i} }: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$

Lemma 2

$\pr_i {\restriction_{Y_i} } = p_i$

$\Box$

Thus $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism.

$\blacksquare$

Proof 2

Let $z \in X$.

Let $i \in I$.

Let $Y_i = \set {x \in X: \forall j \in I \setminus \set i: x_j = z_j}$.

Let $\upsilon_i$ be the subspace topology of $Y_i$ relative to $\tau$.

Let $p_i = \pr_i {\restriction_{Y_i}}$.

Note that by definitions of a restriction and a projection then:

$\forall y \in Y_i: \map {p_i} y = y_i$

$p_i$ is an injection

Let $x, y \in Y_i$.

Then for all $j \in I \setminus \set i$:

$x_j = z_j = y_j$

Let $\map {p_i} x = \map {p_i} y$.

Then:

$x_i = y_i$

Thus:

$x = y$

It follows that $p_i$ is an injection by definition.

$\Box$

$p_i$ is a surjection

Let $x \in X_i$.

Let $y \in Y_i$ be defined by:

$\forall j \in I: y_j = \begin{cases} z_j & j \ne i \\ x & j = i \end{cases}$

Then:

$\map {p_i} y = y_i = x$

It follows that $p_i$ is a surjection by definition.

$\Box$

$p_i$ is a continuous mapping

Let $V \in \tau_i$.

Let $\ds U = \prod_{i \mathop \in I} U_i$ where:

$U_j = \begin{cases} X_j & j \ne i \\ V & j = i \end{cases}$

From Natural Basis of Product Topology, $U$ is an element of the the natural basis.

By definition of the product topology $\tau$ on the product space $\struct {X, \tau}$ the natural basis is a basis for the product topology.

It follows that:

$U$ is open in $\struct {X, \tau}$

Let $x \in Y_i$.

Now:

\(\ds x\)

\(\in\)

\(\ds \map {p_i^\gets} V\)

\(\ds \leadstoandfrom \ \ \)

\(\ds \map {p_i} x\)

\(\in\)

\(\ds V\)

Definition of Inverse Image Mapping of Mapping: $p_i^\gets$

\(\ds \leadstoandfrom \ \ \)

\(\ds \map {\pr_i} x\)

\(\in\)

\(\ds V\)

Definition of Restriction of Mapping $p_i$

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds U\)

Definition of $U$

\(\ds \leadstoandfrom \ \ \)

\(\ds x\)

\(\in\)

\(\ds U \cap Y_i\)

as $x \in Y_i$

By set equality:

$\map {p_i^\gets} V = U \cap Y_i$

By definition of the subspace topology on $Y_i$:

$\map {p_i^\gets} V \in \upsilon_i$

It follows that $p_i$ is continuous by definition.

$\Box$

$p_i$ is an open mapping

Let $U \in \upsilon_i$.

Let $x \in \map {p_i^\to} U$.

Then by definition of the direct image mapping:

$\exists y \in U : x = \map {p_i} y$

By the definition of the subspace topology:

$\exists U' \in \tau: U = U' \cap Y_i$

For all $k \in I$ let $\pr_k$ denote the projection from $X$ to $X_k$.

By definition of the natural basis of the product topology $\tau$:

there exists a finite subset $J$ of $I$

and:

for each $k \in J$, there exists a $V_k \in \tau_k$

such that:

$\ds y \in \bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} \subseteq U'$

Then:

$\ds y \in \paren {\bigcap_{k \mathop \in J} \map{\pr_k^\gets} {V_k} } \cap Y_i \subseteq U' \cap Y_i = U$

By definition of direct image mapping:

$\ds x = \map {p_i} y \in \map {p_i^\to} {\paren {\bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} } \cap Y_i} \subseteq \map {p_i^\to} U$

Recall that $p_i$ is an injection.

Then:

\(\ds \map {p_i^\to} {\paren {\bigcap_{k \mathop \in J} \map {\pr_k^\gets} {V_k} } \cap Y_i}\)

\(=\)

\(\ds \map {p_i^\to} {\bigcap_{k \mathop \in J} \paren {\map {\pr_k^\gets} {V_k} \cap Y_i} }\)

Set Intersection is Self-Distributive

\(\ds \)

\(=\)

\(\ds \bigcap_{k \mathop \in J} \map {p_i^\to} {\map {\pr_k^\gets } {V_k} \cap Y_i}\)

Image of Intersection under Injection

Let $k \in J$.

Lemma

$\map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}$ is open in $\struct{X_i, \tau_i}$

$\Box$

By Open Set Axiom $\paren {\text O 1 }$: Union of Open Sets:

$\ds \bigcap_{k \mathop \in J} \map {p_i^\to} {\map{\pr_k^\gets} {V_k} \cap Y_i}$ is open in $\struct {X_i, \tau_i}$

Since $x \in \map {p_i^\to} U$ was arbitrary then it has been shown that:

$\forall x \in \map {p_i^\to} U : \exists V \in \tau_i : x \in V \subseteq \map {p_i^\to} U$

Then by definition: $\forall x \in \map {p_i^\to} U: \map {p_i^\to} U$ is a neighborhood of $x$.

From Set is Open iff Neighborhood of all its Points:

$\map {p_i^\to} U$ is open in $\tau_i$.

Since $U \in \upsilon_i$ was arbitrary, $p_i$ is an open mapping.

$\blacksquare$