Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Open Mapping

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

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 $p_i = \pr_i {\restriction_{Y_i}}$, where $\pr_i$ is the projection from $X$ to $X_i$.

Then:
 * $p_i$ is an open mapping.

Proof
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 Tychonoff 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:
 * $\displaystyle y \in \bigcap_{k \in J} \map{\pr_k^\gets} {V_k} \subseteq U'$

Then:
 * $\displaystyle y \in \paren{\bigcap_{k \in J} \map{\pr_k^\gets} {V_k}} \cap Y_i \subseteq U' \cap Y_i = U$

By definition of direct image mapping:
 * $\displaystyle x = \map {p_i} y \in \map {p_i^\to} {\paren{\bigcap_{k \in J} \map{\pr_k^\gets} {V_k}} \cap Y_i} \subseteq \map {p_i^\to} U$

Recall that $p_i$ is an injection.

Then:

Let $k \in J$.

Lemma
By open set axiom $O2$, then:
 * $\bigcap_{k \in I_y} \map {p_i^\to} {\map{\pr_k^\gets } {V_k} \cap Y_i}$ is open in $\struct{X_i, \tau_i}$

Since $x \in \map {p_i^\to} U$ was arbitrary then it has been shown that:
 * $\forall x \in \map {p_i^\to} U : \exists V_x \in \tau_i : x \in V_x \subseteq \map {p_i^\to} U$

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

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