Joins of Homeomorphic Spaces are Homeomorphic

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Theorem

Let $X, Y, X', Y'$ be topological spaces.

Suppose that:

$X \sim X'$
$Y \sim Y'$

where $\sim$ denotes homeomorphic spaces.


Then:

$(X \ast Y) \sim (X' \ast Y')$

where $\ast$ denotes the join.


Proof

By definition of homeomorphic, let:

$\phi_X : X \to X'$
$\phi_Y : Y \to Y'$

be homeomorphisms.

Additionally, let:

$I_{\closedint 0 1} : \closedint 0 1 \to \closedint 0 1$

be the identity mapping.

By Identity Mapping is Homeomorphism, we have that $I_{\closedint 0 1}$ is a homeomorphism.


Let $\phi : X \times Y \times \closedint 0 1 \to X' \times Y' \times \closedint 0 1$ be defined as:

$\map \phi {x, y, t} = \tuple {\map {\phi_X} x, \map {\phi_Y} y, \map {I_{\closedint 0 1} } t}$

By Products of Homeomorphic Spaces are Homeomorphic:

$\phi$ is a homeomorphism.


Consider the equivalence relations:

$\RR \subseteq \paren {X \times Y \times \closedint 0 1}^2$
$\RR' \subseteq \paren {X' \times Y' \times \closedint 0 1}^2$

each induced by the mapping:

$\map R {x, y, t} = \begin{cases} x & : t = 0 \\ \tuple {x, y, t} & : t \in \openint 0 1 \\ y & : t = 1 \end{cases}$

which is taken over the corresponding domain and codomain as in the definition of join.


By Quotients of Homeomorphic Spaces are Homeomorphic, the result will follow if we can show:

$\forall a, b \in X \times Y \times \closedint 0 1: \map \RR {a, b} \iff \map {\RR'} {\map \phi a, \map \phi b}$

By definition of the induced relation, it suffices to show that:

$\map R {x, y, t} = \map R {x', y', t'} \iff \map R {\map {\phi_X} x, \map {\phi_Y} y, t} = \map R {\map {\phi_X} {x'}, \map {\phi_Y} {y'}, t'}$

We have:

\(\ds \map R {x, y, t} = \map R {x', y', t'}\) \(\iff\) \(\ds \paren {x = x' \land t = t' = 0}\)
\(\ds \) \(\) \(\, \ds \lor \, \) \(\ds \paren {x = x' \land y = y' \land t = t' \in \openint 0 1}\)
\(\ds \) \(\) \(\, \ds \lor \, \) \(\ds \paren {y = y' \land t = t' = 1}\) Definition of $R$
\(\ds \) \(\iff\) \(\ds \paren {\map {\phi_X} x = \map {\phi_X} {x'} \land t = t' = 0}\)
\(\ds \) \(\) \(\, \ds \lor \, \) \(\ds \paren {\map {\phi_X} x = \map {\phi_X} {x'} \land \map {\phi_Y} y = \map {\phi_Y} {y'} \land t = t' \in \openint 0 1}\)
\(\ds \) \(\) \(\, \ds \lor \, \) \(\ds \paren {\map {\phi_Y} y = \map {\phi_Y} {y'} \land t = t' = 1}\) as $\phi_X$ and $\phi_Y$ are bijections
\(\ds \) \(\iff\) \(\ds \map R {\map {\phi_X} x, \map {\phi_Y} y, t} = \map R {\map {\phi_X} {x'}, \map {\phi_Y} {y'}, t'}\)


Therefore, by the remarks above:


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$\paren {X \ast Y} \sim \paren {X' \ast Y'}$

$\blacksquare$

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