Smoothly homeomorphic for invariance of domain and invariance of dimension

Based on this question Viewing invariance of domain as a converse of invariance of dimension, why then exactly is the following wrong (I mean, I guess it's true for Euclidean spaces $$M$$ and $$N$$ but wrong arbitrary manifolds $$M$$ and $$N$$. I'm just not sure which direction/s, and why)?

Let $$U$$ be an open subset of a smooth $$n$$-manifold $$N$$. Let $$S$$ be a subset of a smooth $$m$$-manifold $$M$$. Let $$U$$ be smoothly homeomorphic to $$S$$. Then $$m=n$$ if and only if $$S$$ is open in $$M$$.

Update: It's true and true even without smoothly. See answer below.

Actually, I think I have a proof of this. I don't even think the smoothness of the homeomorphism is even required. It seems like this is a purely topological result.

Let me know if there are any mistakes.

Theorem. Let $$U$$ be an open nonempty subset of a smooth $$n$$-manifold $$N$$. Let $$S$$ be a subset of a smooth $$m$$-manifold $$M$$. Suppose there is a homeomorphism $$f:U\rightarrow S$$. Then $$m=n$$ if and only if $$S$$ is open in $$M$$.

Proof. $$(\impliedby).$$ Suppose $$S$$ is open in $$M$$. Let $$p\in S$$ (which exists because $$U$$ is nonempty and $$S = f(U)$$), let $$(X, \phi)$$ be a chart of $$M$$ containing point $$p\in S$$, and let $$(Y, \psi)$$ be a chart of $$N$$ containing point $$f^{-1}(p)\in U$$.

We claim $$f(Y\cap U)\cap (X\cap S)$$ is open in $$X$$. Clearly $$Y\cap U$$ is an open subset of $$U$$, so by the homeomorphic property of $$f$$, the set $$f(Y\cap U)$$ is an open subset of $$S = f(U)$$. Since $$S$$ is open in $$M$$, the set $$f(Y\cap U)$$ is open in $$M$$. Since $$S$$ is open, $$f(Y\cap U)\cap S$$ is open in $$M$$. Therefore, $$f(Y\cap U)\cap (X\cap S)$$ is open in $$X$$.

Applying the map $$\phi:X\rightarrow\phi(X)$$ shows $$\phi(f(Y\cap U)\cap (X\cap S))$$ is open in $$\phi(X)\subseteq \mathbb{R}^{m}$$ and hence in $$\mathbb{R}^{m}$$ (because $$\phi(X)$$ is open).

By the similar reasoning as in the above two paragraphs, one can show that the set $$\psi((Y\cap U)\cap f^{-1}(X\cap S))$$ is open in $$\mathbb{R}^{n}$$ (go over the same reasoning as above; I think you need to do this before you move on to the next paragraph below and I don't think there are shortcuts to this).

Now the composition of homeomorphisms $$\psi\circ f^{-1}\circ \phi^{-1}$$ sends $$\phi(f(Y\cap U)\cap (X\cap S))$$ to $$\psi((Y\cap U)\cap f^{-1}(X\cap S))$$. As we have shown, the former is open in $$\mathbb{R}^{m}$$ and the latter is open in $$\mathbb{R}^{n}$$. Also, they are nonempty because the former contains $$\phi(p)$$ and the latter contains $$\psi(f^{-1}(p))$$. By applying the last theorem I wrote in this post, we deduce that $$m=n$$.

$$(\implies).$$ Suppose $$m=n$$. Let $$p\in S$$. Let $$(X, \phi)$$ be a chart for $$M$$ containing $$p$$, and let $$(Y, \psi)$$ be a chart for $$N$$ containing $$f^{-1}(p)$$.

By the homeomorphism property, $$f(Y\cap U)$$ is open in $$S = f(U)$$, and by the subspace topology, $$f(Y\cap U)\cap (X\cap S)$$ is also open in $$S$$. By applying $$f^{-1}$$, the set $$(Y\cap U)\cap f^{-1}(X\cap S)$$ is open in $$U$$. Since $$U$$ is open, that set is open in $$M$$; since $$Y$$ is open, that set is open in $$Y$$.
By applying $$\psi:Y\rightarrow\psi(Y)$$, the set $$\psi((Y\cap U)\cap f^{-1}(X\cap S))$$ is open in $$\psi(Y)$$, which means it is open in $$\mathbb{R}^{n}$$.

Then $$\phi\circ f\circ\psi^{-1}:\psi((Y\cap U)\cap f^{-1}(X\cap S))\rightarrow \phi(f(Y\cap U)\cap (X\cap S))\subseteq\mathbb{R}^{n}$$ is a continuous injective map, and we know the domain $$\psi((Y\cap U)\cap f^{-1}(X\cap S))$$ is open in $$\mathbb{R}^{n}$$ (we can't say the same about the image of the function, however, because not knowing $$S$$ is open means we don't know whether $$X\cap S$$ is open in $$X$$). By applying Theorem 2 of this link, we find that the set $$\phi(f(Y\cap U)\cap (X\cap S))$$ is open in $$\mathbb{R}^{n}$$. Then $$f(Y\cap U)\cap (X\cap S)$$ is open in $$X$$, and hence in $$M$$ (because $$X$$ is itself open).

But now we've shown something interesting: for any $$p\in S$$, there exists an open subset of $$M$$ of the form $$O_{p} = f(Y\cap U)\cap (X\cap S)$$ such that $$p\in O_{p}\subseteq S.$$ Therefore, we can write $$S$$ as a union of open sets: $$S = \bigcup_{p\in S} O_{p}$$ and therefore $$S$$ is open in $$M$$. $$\tag*{\blacksquare}$$

• Oh this is great. Thanks a lot SpiralRain! So then smoothness/diffeomorphicness just makes it possible to think of tangent spaces and simplify the proof but they're actually not necessary for (both invariance of dimension, well this one I already know, and) invariance of domain?
– user636532
Jul 21 '19 at 5:55
• @SeleneAuckland Apparently yes! Jul 21 '19 at 5:58