# Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Lemma 1

## 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 $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.

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

Let $z \in X$.

Let $i, k \in I$.

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

Let $p_i = \pr_i {\restriction_{Y_i} }$ be the restriction of $\pr_i$ to $Y_i$.

Let $V_k \in \tau_k$.

Let $\map {\pr_k^\gets } {V_k} \cap Y_i \ne \O$.

Then:

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

## Proof

We have that $p_i$ is a bijection from the lemmas:

$p_i$ is an injection
$p_i$ is a surjection

Let $x \in X_i$.

Then:

 $\ds x$ $\in$ $\ds \map {p_i^\to} {\map {\pr_k^\gets} {V_k} \cap Y_i}$ $\ds \leadstoandfrom \ \$ $\ds \map {p_i^{-1} } x$ $\in$ $\ds \map {\pr_k^\gets} {V_k} \cap Y_i$ Definition of Direct Image Mapping of Mapping $\ds \leadstoandfrom \ \$ $\ds \map {p_i^{-1} } x$ $\in$ $\ds \map {\pr_k^\gets} {V_k}$ as $\map {p_i^{-1} } x \in Y_i$ $\ds \leadstoandfrom \ \$ $\ds \map {\pr_k} {\map {p_i^{-1} } x}$ $\in$ $\ds V_k$ Definition of Direct Image Mapping of Mapping

By definition of $p_i$:

$\map {p_i^{-1} } x = y$

where:

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

#### Case 1: $k = i$

Let $k = i$.

Then:

 $\ds \map {\pr_i} {\map {p_i^{-1} } x}$ $=$ $\ds \map {\pr_i} {y}$ $\ds$ $=$ $\ds y_i$ $\ds$ $=$ $\ds x$

So:

$x \in \map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i}$ if and only if $x \in V_i$

That is:

$\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = V_i \in \tau_i$

$\Box$

#### Case 2: $k \ne i$

Let $k \ne i$.

Then:

 $\ds \map {\pr_k} {\map {p_i^{-1} } x}$ $=$ $\ds \map {\pr_k} y$ $\ds$ $=$ $\ds y_k$ $\ds$ $=$ $\ds z_k$

So:

$x \in \map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i}$ if and only if $z_k \in V_k$

Since $x \in X_i$ was arbitrary, then:

$\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = X_i$ if and only if $z_k \in V_k$

Let $w \in \map {\pr_k^\gets} {V_k} \cap Y_i$ which is guaranteed since $\map {\pr_k^\gets} {V_k} \cap Y_i \ne \O$

By the definition of inverse image mapping then:

$w_k = \map {\pr_k} w \in V_k$

By the definition of $Y_i$ then:

$w_k = z_k$

So $z_k \in V_k$.

It follows that:

$\map {p_i} {\map {\pr_i^\gets} {V_i} \cap Y_i} = X_i \in \tau_i$

$\Box$

In either case:

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

$\blacksquare$