Subspace of Product Space is Homeomorphic to Factor Space/Proof 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 $\ds \struct {X, \tau} = \prod_{i \mathop \in I} \struct {X_i, \tau_i}$ be the product space of $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$.

Suppose that $X$ is non-empty.

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

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 $Y_i$.

Proof

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 $\tau$.

For all $j \in I$ let:

$Z_j = \begin{cases} X_i & i = j \\

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

Lemma 1

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

$\Box$

From Product Space of Subspaces is Subspace of Product Space , $\struct {Y_i, \upsilon_i}$ is a product space.

Consider the projection:

$p_i: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$

From Projection from Product of Family is Injection iff Other Factors are Singletons, $p_i$ is injective.

From Projection from Product of Family is Surjective, $p_i$ is surjective.

From Projection from Product Topology is Continuous:General Result, $p_i$ is continuous.

From Projection from Product Topology is Open:General Result, $p_i$ is open.

Thus, by definition, we have that $p_i$ is a homeomorphism.

Consider the projection:

$\pr_i: \struct {X, \tau} \to \struct {X_i, \tau_i}$

and the restriction:

$\pr_i {\restriction_{Y_i} }: \struct {Y_i, \upsilon_i} \to \struct {X_i, \tau_i}$

Lemma 2

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

$\Box$

Thus $\pr_i {\restriction_{Y_i} }: Y_i \to X_i$ is a homeomorphism.

$\blacksquare$