Subspace of Product Space is Homeomorphic to Factor Space/Proof 2
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 $Y_i$.
Proof
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 an 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 $\tau$:
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.
$\blacksquare$