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Question: Let $X,Y$ be topological spaces, $x \in X, y \in Y$, and let $\mathscr{N}_x, \mathscr{N}_y$ denote the neighborhood systems of $x$ in $X$ and $y$ in $Y$ respectively. Then is there a criterion to say that $\mathscr{N}_x \cong \mathscr{N}_y$?

Any suggestions or pointers to references which explain or discuss such a criterion (or criteria, or multiple notions of equivalence, etc.) would be sufficient for an answer.

Example: Let $G$ be a topological group, then "intuitively" we should have for all $g \in G$ that $\mathscr{N}_g \cong \mathscr{N}_e$ using translation. In particular, $G$ is first countable if and only if the neighborhood system of the identity $\mathscr{N}_e$ has a countable basis.

Attempt: Analogous to the notion of homeomorphism, it occurs to me to try to use the notion of continuity at a point. One says that a function $f: X \to Y$ is continuous at $x \in X$ iff, for every neighborhood of $f(x) \in Y$, the preimage under $f$ contains a neighborhood of $x$. (See also.)

Now let $f: X \to Y, g: Y \to X$, not necessarily continuous, not necessarily sections/retractions/inverses, such that $f(x)=y$ and $g(y)=x$. Then would it make sense to say that $\mathscr{N}_x \cong \mathscr{N}_y$ if and only if there exist such $f,g$ with $f$ continuous at $x$ and $g$ continuous at $y$?

Such a relation is obviously reflexive and symmetric, transitivity follows from the claim:

If $h: X \to Y$ is continuous at $x \in X$ and $\ell: Y \to Z$ is continuous at $h(x)$, then $\ell \circ h: X \to Z$ is continuous at $x$.

Note: Perhaps the definition should be like that of germs-- I think that such a similarity might follow from the above, but I'm not sure.

Chill2Macht
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    $f$ is continuous at $x \in X$ iff for every neighbourhood of $f(x) \in Y$ the preimage under $f$ is a neighbourhood of $x$. Or, if your definition of neighbourhood says neighbourhoods are open, then it's "contains a neighbourhood". Being contained in a neighbourhood isn't a useful criterion, $X$ is a neighbourhood of $x\in X$. And constant functions ought to be continuous. – Daniel Fischer Jan 22 '17 at 12:51
  • @DanielFischer I fixed the mistake in the post. As you guessed, I am using the definition of neighborhood that says neighborhoods are open. Another way to see that you are right is to consider the special case of metric spaces: what I wrote does not correspond to "for every $\varepsilon>0$ there exists a $\delta > 0$ such that $|x_0 - x|< \delta \implies |f(x_0) - f(y)| < \varepsilon$", whereas your definition does. – Chill2Macht Jan 22 '17 at 14:42
  • Looking at this now the answer is probably something related to "local homeomorphism" or "local homeomorphism at a point" https://en.wikipedia.org/wiki/Local_homeomorphism – Chill2Macht Mar 06 '22 at 03:11

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