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S Jan 23 at 14:24 review First answers
Jan 23 at 14:50
S Jan 23 at 14:24 history edited user515519 CC BY-SA 4.0
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Jan 23 at 14:23 comment added user515519 @MoisheKohan if $\varphi:U\to X$ is a diffeomorphism then for $x\in X$ lying in the closure of $U$ you cannot define its image since a sequence $(x_n)$ of points in $U$ converging to $x$ the image $(\varphi(x_n))$ converges to infinity.
Jan 23 at 14:17 comment added Moishe Kohan Why would there be "no way"?
S Jan 23 at 13:48 review First answers
Jan 23 at 14:11
S Jan 23 at 13:48 history answered user515519 CC BY-SA 4.0