Topological Product with Singleton

Theorem
Let $T_1$ and $T_2$ be topological spaces.

Let $a \in T_1, b \in T_2$.

Let $T_1 \times T_2$ be the topological product of $T_1$ and $T_2$.

Then:
 * $T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$
 * $T_2$ is homeomorphic to the subspace $\left\{{a}\right\} \times T_2$ of $T_1 \times T_2$

Proof
The conclusions are symmetrical, so we will show that $T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$

Let $f\colon T_1 \to T_1 \times \left\{{b}\right\}$ with $f\left({x}\right) = (x,b)$.

$f$ is clearly a bijection.

$f^{-1}$ is a restriction to the subspace of the projection of $T_1 \times T_2$ onto $T_1$, which is continuous, so $f^{-1}$ is continuous.

$f$ is continuous:

Let $x \in T_1$, and let $U$ be an open set in $T_1 \times \left\{{b}\right\}$ such that $f\left({x}\right) \in U$.

Then for some $U'$ open in $T_1 \times T_2$, $U' \cap \left({T_1 \times \left\{{b}\right\}}\right) = U$.

By the definition of the product topology, there are open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that $f\left({x}\right) = \left({x,b}\right) \in V_1 \times V_2 \subseteq U'$.

Then for any $y \in V_1$, $f\left({y}\right)=\left({y,b}\right) \in V_1 \times V_2 \subseteq U'$,

But $\left({y,b}\right) \in T_1 \times \left\{{b}\right\}$, so $\left({y,b}\right) \in U$.

Thus if $y \in V_1$, $f\left({y}\right) \in U$.

A Generalization
Let $X_a$ be a topological space for each $a \in I$ and let $\displaystyle X = \prod_{a \mathop\in I} X_a$. Let $p_a$ be the $a$th projection for each $a \in I$. Suppose that $x_0 \in X$ and $k \in I$.

Let $S = \left\{{x \in X: p_a(x) = p_a\left({x_0}\right) \text{ when $a≠k$}}\right\}$

Then $S$ is homeomorphic to $X_k$