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 $\displaystyle \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ 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:

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:

So:
 * $x \in \map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i}$ $x \in V_i$

That is:
 * $\map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i} = V_i \in \tau_i$

Case 2: $k \neq i$
Let $k \ne i$.

Then:

So:
 * $x \in \map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i}$ $z_k \in V_k$

Since $x \in X_i$ was arbitrary, then:
 * $\map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i} = X_i$ $z_k \in V_k$

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

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