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Let $M=(0,1) \times (0,1)$. Instead of the usual topology, let us define a basis of neighbourhood of the point $(x,y)$ as the set of all open intervals $I_\delta(x,y)=\{x\} \times (y-\delta, y+\delta)$ where $0 < \delta < \min\{y, 1-y\}$. Define local charts near $(x,y)$ by the maps $I_\delta(x,y) \to \mathbb R: (x,t) \mapsto t$

  1. Am I mistaken for thinking that $M$, endowed with these local charts, is a dimension $1$ manifold, say $M_1$?

  2. Now consider $M$ endowed with the usual topology. It is a dimension $2$ manifold, say $M_2$. Am I mistaken for thinking that the identity of $M$ is an immersion $M_1 \to M_2$?

Régis
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  • What is your definition of a manifold? Does it include the 2nd countable condition? – Moishe Kohan Apr 04 '25 at 18:16
  • Cf. this question. – Moishe Kohan Apr 04 '25 at 18:22
  • If I am not mistaken the reference I am using does not require 2nd countable condition. I'd have to check, though, as I don't have it under my eyes right now. – Régis Apr 04 '25 at 19:06
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    Some sources only assume paracompactness and this holds in your example. But you should also think of a more general example: Take any manifold $M$ and consider the same set with discrete topology, $M_1$. You get a bijective immersion $M_1\to M$. – Moishe Kohan Apr 04 '25 at 19:39
  • Ok I checked and the source I am using only assumes Hausdorff space. – Régis Apr 07 '25 at 07:21

1 Answers1

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What an interesting topology! Your $M_1$ is, in essence, the disjoint union of uncountably many copies of $\mathbb{R}$. So although $M_1$ is locally Euclidean and Hausdorff, it lacks 2nd countability.

As for the map $M_1 \rightarrow M_2$, this is certainly an immersion. Locally, from the perspective of a point in $M_1$, the inclusion into $M_2$ just looks like the inclusion of an interval contained in $M_2$.

JMM
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