Subspace of Product Space is Homeomorphic to Factor Space/Product with Singleton/Proof 2

Proof
The conclusions are symmetrical.

, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \left\{{b}\right\}$ of $T_1 \times T_2$.

Let $f: T_1 \to T_1 \times \set{b}$ be defined as:
 * $\map f x = \tuple {x, b}$

Lemma
$f^{-1}$ is a restriction to the subspace $T_1 \times \set {b}$ of the projection $\pr_1$ of $T_1 \times T_2$ onto $T_1$.

From Projection from Product Topology is Continuous, $\pr_1$ is a continuous.

From Restriction of Continuous Mapping is Continuous, $f^{-1}$ is a continuous.

It is to be shown that $f$ is continuous.

Let $x \in T_1$.

Let $U$ be an open set in $T_1 \times \set {b}$ such that:
 * $\map f x \in U$

By definition of the subspace topology then for some $U'$ open in $T_1 \times T_2$:
 * $U' \cap \paren {T_1 \times \set {b}}= U$

By the definition of the Natural Basis of Tychonoff Topology, there exist open sets $V_1$ and $V_2$ in $T_1$ and $T_2$, respectively, such that:
 * $\map f x = \tuple {x, b} \in V_1 \times V_2 \subseteq U'$

Then for any $y \in V_1$:
 * $\map f y = \tuple {y, b} \in V_1 \times V_2 \subseteq U'$

But:
 * $\tuple {y,b} \in T_1 \times \set{b}$

so:
 * $\tuple{y, b} \in U$

Thus if $y \in V_1$, it follows that $\map f y \in U$.

Then $f$ is continuous by definition.