Homeomorphism Relation is Equivalence

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Theorem

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

Let $T_1 \sim T_2$ denote that $T_1$ and $T_2$ are homeomorphic.


The relation $\sim$ is an equivalence relation.


Proof

Checking in turn each of the criteria for equivalence:


Reflexivity

Let $T$ be a topological space.

From Identity Mapping is Homeomorphism, the identity mapping $I_T: T \to T$ is a homeomorphism.

So $T \sim T$, and $\sim$ has been shown to be reflexive.

$\Box$


Symmetry

Let $T_1$ and $T_2$ be topological spaces such that $T_1 \sim T_2$.

By definition, there exists a homeomorphism $f: T_1 \to T_2$.

From Inverse of Homeomorphism is Homeomorphism it follows that $f^{-1}: T_2 \to T_1$ is also a homeomorphism.

So $T_2 \sim T_1$, and $\sim$ has been shown to be symmetric.

$\Box$


Transitivity

Let $T_1, T_2, T_3$ be topological spaces such that $T_1 \sim T_2$ and $T_2 \sim T_3$.

By definition, there exist homeomorphisms $f: T_1 \to T_2$ and $g: T_2 \to T_3$.

From Composite of Homeomorphisms is Homeomorphism it follows that $g \circ f: T_1 \to T_3$ is also a homeomorphism.

So $T_1 \sim T_3$, and $\sim$ has been shown to be transitive.

$\Box$


$\sim$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$


Sources

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