# Smooth invariance of domain and diffeomorphisms

A question regarding Tu's take on smooth invariance of domain. The theorem is stated as follows:

Let $$U \subset \mathbb{R}^n$$ be an open subset, $$S \subset \mathbb{R}^n$$ an arbitrary subset, and $$f:U \rightarrow S$$ a diffeomorphism. Then S is open in $$\mathbb{R}^n$$.

I understand the subtlety of proving that $$S$$ is open in $$\mathbb{R}^n$$, as it is only given that $$S$$ is open in $$S$$ itself. Roughly speaking, the proof is as follows. For arbitrary $$p \in U$$ and $$f(p) \in S$$, there is guaranteed to be an open set $$V$$ in $$\mathbb{R}^n$$ with a $$C^\infty$$ map $$g:V \rightarrow \mathbb{R}^n$$ such that $$g|_S = f^{-1}$$. Via chain rule, the pushforward $$f_{*,p}:T_pU \rightarrow T_{f(p)}V$$ is invertible, and therefore an isomorphism, which gurantees that $$f$$ is a local diffeomorphism at $$p$$ from the inverse function theorem. From there, it can be shown that $$S$$ is open in $$\mathbb{R}^n$$ via local criterion for openness after a quick hop, skip, and a jump.

... However, perhaps this is a silly question, but hasn't the conclusion of this proof already been assumed from the beginning, simply by stating that $$f:U \rightarrow S$$ is a diffeomorphism, because S must be a manifold for the definition of a diffeomorphism to hold: diffeomorphisms map only between manifolds, right? And if $$S$$ is a subset of $$\mathbb{R}^n$$, it must be open in $$\mathbb{R}^n$$, because open sets of manifolds are also manifolds, and a closed/non-open subset of $$\mathbb{R}^n$$ can't be a manifold, yeah? Sure, a non-open subset $$S$$ can be given the subset topology, but if $$S$$ isn't open in $$\mathbb{R}^n$$, then how can we force a differentiable structure on it to make it a manifold in the first place? It sure doesn't seem like I can force charts/homeomorphisms to $$\mathbb{R}^n$$ to hold well when the ambient space is a non-open subset of $$\mathbb{R}^n$$.

To repeat: I know that $$S$$ is only assumed to be open in $$S$$ itself at the start, but the definition of diffeomorphism as a map between manifolds doesn't even make sense if $$S$$ isn't a manifold, and if $$S$$ is a manifold and a subset of $$\mathbb{R}^n$$, it has to be open in $$\mathbb{R}^n$$, because non-open subsets of $$\mathbb{R}^n$$ can't be given manifold structure. Isn't the proof then then just proving something that was assumed by definition?

• What is the definition of a diffeomorphism between not necessarily open subsets of $\mathbb R^n$? Commented Apr 5, 2021 at 23:08
• Ah, indeed. For a non-open set, it's smooth at all points in it as long as an open nbh $U$ in $\mathbb{R}^n$ with $g:U \rightarrow \mathbb{R}^n$ such that $g = f$ on the restriction to the overlap between $U$ and $S$. That's what I get for taking too long a break! Apologies! Commented Apr 6, 2021 at 4:41

Perhaps Tu's presentation of basic concepts could be improved didactically. In his book he defines

• smooth maps $$f : U \to \mathbb R^m$$, where $$U \subset \mathbb R^n$$ is open - but he does not explicitly introduce the concept of as smooth map $$f : U \to V$$ for an open $$V \subset \mathbb R^m$$. Okay, I would not regard that as real gap.

• diffeomomorphisms $$F : U \to V$$ between open subsets $$U, V \subset \mathbb R^n$$.

• smooth maps between manifolds

• diffeomorphism between manifolds

Then, in Definition 22.1 he suddenly defines

• smooth maps $$f : S \to \mathbb R^m$$ for arbitrary subsets $$S \subset \mathbb R^n$$

After this he remarks

With this definition it now makes sense to speak of an arbitrary subset $$S \subset \mathbb R^n$$ being diffeomorphic to an arbitrary subset $$T \subset \mathbb R^m$$; this will be the case if and only if there are smooth maps $$f : S \to T \subset \mathbb R^m$$ and $$g : T \to S \subset \mathbb R^n$$ that are inverse to each other.

• He does, my fault. I seemed to have glossed over the expanded definition, mistakenly only referring the definition of diffeomorphism presented for manifolds in chapter 5. Thanks much! Commented Apr 6, 2021 at 21:26