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Let $g:U \subset {\mathbb R}^n \rightarrow V \subset {\mathbb R}^n$ a diffeomorphism. Is the restriction of $g$ to any subset of $U$ still a diffeomorphism onto its image?

I think that it is still a diffeomorphism, because it should be a differentiable function by the definition. But it is a bit strange if the subset of $U$ is for exemple a single point.

DuduBob
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Johny06
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2 Answers2

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The restriction to any open subset $M \subset U$ is still a diffeo onto the image $g(M) \subset U$, as you should be able to see from the definition.

More generally, if the subset $M$, when equipped with the topology and differentiable structure induced from $\mathbb{R}$, is a differentiable manifold, then $g: M \to g(M)$ will be a diffeo.

For weirder subsets it may not make sense to talk about diffeos, since these are basically only defined on differential manifolds. For instance, if $g$ is originally a diffeo from $\mathbb{R}^2$ onto itself and you consider the subset $M$ to be two intersecting lines, it makes no sense to talk about diffeos on $M$, because it has no differentiable structure of its own (one can not define a chart around the intersection point).

Of course, in the situation you describe it will always be possible to extend $g\rvert_M$ to a diffeo on a larger space, so the function $g\rvert_M$ will have nice properties, but it may itself not be a diffeo because of technical reasons, as exemplified above.

usbc
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    Nice answer! Maybe you want to stress that you need your manifold to be a differentiable one, so, it should not only inherit its topology from $\mathbb{R}^n$. – Severin Schraven Sep 09 '22 at 16:01
  • $g(M)$ is also a manifold with the induced topology if $M$ is a manifold ? – Johny06 Sep 09 '22 at 16:13
  • If you come back here, i have an other question, i'm a bit confused on somethings https://math.stackexchange.com/questions/4528117/question-about-real-submanifolds – Johny06 Sep 09 '22 at 17:16
  • Some authors (for example Tu, Loring W. An Introduction to Manifolds. Springer, New York, NY, 2011) define smooth maps $f : S \to \mathbb R^n$ on arbitrary subsets $S \subset \mathbb R^m$ by requiring that they are restrictions of smooth maps $F : U\to \mathbb R^n$ with an open $U \subset \mathbb R^n$ such that $S \subset U$. See also https://math.stackexchange.com/q/4090839 and https://math.stackexchange.com/q/3511868. This allows to extend the concept of "diffeomorphism" to arbitrary subsets of Euclidean spaces. – Paul Frost Sep 09 '22 at 17:32
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In general, we define derivatives on opens sets. We do so, to make possible "approach any point from any direction". The only exceptions to this that I know is the $\mathbb R$, where you can define one-sided derivatives and directional derivatives. But they both are something a bit different from regular derivatives.

Diffeomorphism in its definition states that it is differentiable in every point of its domain. This implies that such domain must be an open set.

You can definitely restrict your domain to any open non-empty subset and obtain diffeomorphism. But restricting to arbitrary set gives you for sure only homeomorphism.

Esgeriath
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