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

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

Suppose that $X$ is non-empty.

Then for each $i \in I$ there is a subspace $Y_i \subseteq X$ which is homeomorphic to $T_i = \struct {X_i, \tau_i}$.

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 $Y_i$.

Let $z \in X$.

Let $i \in i$.

For all for all $j \in I$ let:
 * $Z_j = \begin{cases} X_i & i = j \\

\set{z_j} & j \neq i \end{cases}$

Let $Y_i = \prod_{j \mathop \in I} Z_j$

Then:
 * $\pr_i {\restriction_{Y_i} } = \pr'_i$

Proof
For all $y \in Y_i$:

By equality of mappings, $\pr_i {\restriction_{Y_i} } = \pr'_i$