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?

  • 1
    $\begingroup$ What is the definition of a diffeomorphism between not necessarily open subsets of $\mathbb R^n$? $\endgroup$
    – Paul Frost
    Commented Apr 5, 2021 at 23:08
  • $\begingroup$ 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! $\endgroup$
    – Jerome
    Commented Apr 6, 2021 at 4:41

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$
    – Jerome
    Commented Apr 6, 2021 at 21:26

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