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

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Theorem

Let $T_1$ and $T_2$ be non-empty topological spaces.

Let $b \in T_2$.

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

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

Then:

$T_1$ is homeomorphic to the subspace $T_1 \times \set b$ of $T_1 \times T_2$
$T_1$ is homeomorphic to the subspace $\set b \times T_1$ of $T_2 \times T_1$


Proof 1

From Finite Cartesian Product of Non-Empty Sets is Non-Empty both $T_1 \times T_2$ and $T_2 \times T_1$ are both non-empty.

The conclusions follow immediately from Subspace of Product Space is Homeomorphic to Factor Space.

$\blacksquare$


Proof 2

The conclusions are symmetrical.

Without loss of generality, therefore, it will be shown that $T_1$ is homeomorphic to the subspace $T_1 \times \set b$ 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

$f$ is a bijection.

$\Box$


$f^{-1}$ 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 Product 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.

$\blacksquare$


Sources