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

Then:
 * $p_i$ is an open mapping.

Proof
Let $U \in \upsilon_i$.

Then by the definition of the subspace topology:
 * $\exists U' \in \tau: U = U' \cap Y_i$

By definition of the natural basis of the Tychonoff topology $\tau$:
 * for each $y \in U$, there exists a finite subset $I_y$ of $I$

and:
 * for each $k \in I_y$, there exists a $V_k \in \tau_k$

such that:
 * $\displaystyle y \in \bigcap_{k \in I_y} \map{\pr_k^\gets} {V_k} \subseteq U'$

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

Then:
 * $\displaystyle \map {p_i} y \in \map {p_i^\to} {\paren{\bigcap_{k \in I_y} \map{\pr_k^\gets} {V_k}} \cap Y_i} \subseteq \map {p_i^\to} U$

Now:

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 $\struct{X_i, \tau_i}$