Characterization of Homeomorphic Topological Spaces/Necessary Condition
Jump to navigation
Jump to search
![]() | This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Theorem
Let $T_1 = \struct{S_1, \tau_1}$ be topological space.
Let $S_2$ be a set.
Let $\tau_2$ be a subset of the powerset $\powerset {S_2}$.
Let $\struct{S_2, \tau_2}$ be a topological space homeomorphic to $T_1$.
Then:
- there exists a mapping $f : S_1 \to S_2$:
- where
- $f^\to \restriction_{\tau_1}$ denotes the restriction of $f^\to$ to $\tau_1$
- $f^\to$ denotes the direct image mapping of $f$
Proof
Let $f: S_1 \to S_2$ be a homeomorphism.
By definition of a homeomorphism:
- $f$ is a bijection
- $f$ is an open mapping
- $f$ is a continuous mapping
By definition of an open mapping:
- $\forall U \in \tau_1 : f \sqbrk U \in \tau_2$
By definition of direct image mapping:
- $\forall U \in \tau_1 : \map {f^\to} U \in \tau_2$
By definition of continuous mapping:
- $\forall V \in \tau_2 : f^{-1} \sqbrk V \in \tau_1$
We have:
\(\ds \forall V \in \tau_2: \, \) | \(\ds \map {f^\to} {f^{-1} \sqbrk V}\) | \(=\) | \(\ds f \sqbrk {f^{-1} \sqbrk V}\) | Definition of Direct Image Mapping | ||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren{f \circ f^{-1} } } V\) | Definition of Composite Mapping | |||||||||||
\(\ds \) | \(=\) | \(\ds V\) | Image of Preimage of Subset under Surjection equals Subset |
It follows that $f^\to \restriction_{\tau_1}$ is a surjection from $\tau_1$ to $\tau_2$.
$\blacksquare$