Subspace of Product Space is Homeomorphic to Factor Space/Proof 2/Continuous 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 continuous.

Proof
Let $V \in \tau_i$.

Let $\displaystyle U = \prod_{i \mathop \in I} U_i$ where:
 * $U_j = \begin{cases} X_j & j \ne i \\ V & j = i \end{cases}$

From Natural Basis of Tychonoff Topology, $U$ is an element of the the natural basis.

By definition of the Tychonoff topology $\tau$ on the product space $\struct {X, \tau}$ the natural basis is a basis for the Tychonoff topology.

It follows that:
 * $U$ is open in $\struct {X, \tau}$

Let $x \in Y_i$.

Now:

By set equality:
 * $ \map {p_i^\gets} V = U \cap Y_i$

By definition of the subspace topology on $Y_i$:
 * $\map {p_i^\gets} V \in \upsilon_i$

It follows that $p_i$ is continuous by definition.