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

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Theorem

Let $\family {X_i}_{i \mathop \in I}$ be a family of sets where $I$ is an arbitrary index set.

Let $\ds X = \prod_{i \mathop \in I} X_i$ be the Cartesian product of $\family {X_i}_{i \mathop \in I}$.

Let $z \in X$.

Let $i \in I$.

Let $\pr_i : X \to X_i$ be the $i$th-projection from $X$.

For all for all $j \in I$ let:

$Z_j = \begin{cases} X_i & i = j \\ \set{z_j} & j \ne i \end{cases}$

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

Let $p_i : Y_i \to X_i$ be the $i$th-projection from $Y_i$.


Then:

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


Proof

For all $y \in Y_i$:

\(\ds \map {\pr_i {\restriction_{Y_i} } } y\) \(=\) \(\ds \map {\pr_i} y\) Definition of Restriction of Mapping: $\pr_i {\restriction_{Y_i} } : Y_i \to X_i$
\(\ds \) \(=\) \(\ds y_i\) Definition of Projection: $\pr_i: X \to X_i$
\(\ds \) \(=\) \(\ds \map {p_i} y\) Definition of Projection: $p_i: Y_i \to X_i$

By Equality of Mappings:

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

$\blacksquare$