Show that an invertible immersion $\varphi$:N $\to$ M is a diffeomorphism. Give a counterexample to this statement if N does not have a countable basis. I think I can use the Constant Rank Theorem. Since it is invertible, the dimension of N and M should be same.Then we know it is a diffeomorphism, but seems it has nothing to do with a countable basis? Thanks!
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Why do you think there is a counter example without countability? – Moishe Kohan Apr 29 '14 at 02:19
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Cause it asks for the counterexample...It seems if it is not second countable, the dimension of N and M could be different. – user146507 Apr 29 '14 at 05:24
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What is "it" in "it asks"? – Moishe Kohan Apr 29 '14 at 15:22
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Hint: Take $N$ to be ${\mathbb R}$ with discrete topology.
Moishe Kohan
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It is a manifold, but is not locally $R$; think, what is its dimension? – Moishe Kohan Apr 30 '14 at 02:05
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under the usual topology of $R^n$,if the map between R with discrete topology and $R^n$ is identity map, then it is not an open map since a dot is not open. – user146507 Apr 30 '14 at 12:44
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Yes that will be a $0$-dimensional manifold but how is differentiability defined so that we can even talk about immersions ? @MoisheKohan – Someone Sep 27 '20 at 15:27
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@Something: Smoothness for 0-dimensional manifolds was discussed several times at MSE, for instance here, here. The bottom line is that every map of a 0-dimensional manifold to another manifold will be smooth. – Moishe Kohan Sep 28 '20 at 19:01