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I have a question from the answer to this question. It shows that if $f_n: M \rightarrow N$ is a sequence of (Riemannian) isometries that converge uniformly to a map $f: M \rightarrow N$, then $f$ is a Riemannian isometry. To do so, we use the fact that Riemannian isometry and distance isometry coincide.

In the proof, Prof. Lee's hint tells us that it is enough to show $f(M)$ is closed. However, from here how do we conclude that $f(M) = N$? Do we have any knowledge that $f(M)$ is open? Any insights into this will be appreciable.

Arctic Char
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Aniruddha Deshmukh
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  • You should start by defining a Riemannian isometry; from the context, it appears that your definition includes the requirement that it is a diffeomorphism. Then argue that $dim(M)=dim(N)$ and that the differential of each Riemannian isometry in this case is a linear isomorphism at each points. Then use IFT to conclude that each isometry is an open map... – Moishe Kohan Aug 09 '22 at 12:48
  • With the information that is given, we cannot conclude that $f$ is smooth, let alone a local diffeomorphism. The questions asks whether from the knowledge that $f$ is a uniform limit of diffeomorphisms we can conclude that it is a local diffeomorphism (or an open map at the least) to proceed. – Aniruddha Deshmukh Aug 09 '22 at 14:35
  • I am not asking about $f$, I am asking about your definition of an isometry. – Moishe Kohan Aug 09 '22 at 16:01
  • The definition of Riemannian isometry requires it to be a diffeomorphism. But indeed, one can prove that Riemannian isometries and metric isometries (distance preserving homeomoprhisms) are the same. – Aniruddha Deshmukh Aug 10 '22 at 05:39

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