Do homeomorphic smooth manifolds, like diffeomorphic ones, have the same dimension? [duplicate]

This question already has an answer here:

For smooth manifolds $$A$$ and $$B$$ with respective dimensions $$a$$ and $$b$$. If $$A$$ and $$B$$ are diffeomorphic, then $$a=b$$. I guess the same is true for homeomorphic topological ($$C^0$$, I guess) manifolds (with dimension).

1. If $$A$$ and $$B$$ were instead homeomorphic, then would we still have $$a=b$$?

• I imagine some weird case how they would have the same dimension viewing them as topological manifolds (with dimension), but they might have different dimensions as smooth manifolds. I actually haven't encountered anything like different dimensions depending on which level of $$C^k$$ we are since most the manifolds so far I've been studying are smooth, but I think it's similar to the idea of Isomorphic groups but not isomorphic rings.

If yes, then follow-up questions:

1. Does the "$$B$$" in this question What does it take for a smooth homeomorphism to be a diffeomorphism? have dimension $$k$$ because $$\dim B = \dim A = \dim \mathbb R^k = k$$ ?

2. Can Corollaries 8.6 and 8.7 in An Introduction to Manifolds by Loring W. Tu be relaxed as follows?

• 3.1. For 8.6: change "diffeomorphism" to "smooth homeomorphism"

• 3.2. For 8.7: change "diffeomorphic" to simply "homeomorphic"

marked as duplicate by Eric Wofsey general-topology StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 21 at 2:36

• Of course. Homeomorphic smooth manifolds are homeomorphic topological manifolds... – Bach Jul 20 at 7:34
• – Bach Jul 20 at 7:56
• @bach: these are diffeomorphic (but not via the identity). The dimension is a topological invariant, which follows from invariance of domain for example. – Thomas Rot Jul 20 at 7:56
• Your post has too many questions and subquestions and followup questions. This is proved by the long rambling comment thread that follows. Please keep the number of different things that you ask in your post down to a minimum, and you'll get better responses. – Lee Mosher Jul 20 at 13:06
• No...don't do that! There's an implicit deal: you ask a question, someone gives an answer, you mark it correct, they get a few silly internet "points" which makes them feel good. If the answer helps you realize there's something else you don't know, you ask a new question; you don't get to troll them along, making them your math tutor forever. It's just plain rude when you violate these community norms. One thing you can do is say "Thanks; now that I understand a little more, I've posted a followup question." It's also totally fine to ask clarifying questions on an answer, of course. – John Hughes Jul 20 at 14:25

If I'm reading your comments correctly, then you seem to have the right idea. I thought maybe it can help if I write the following below.

A topological manifold has a topological atlas $$\mathcal{A} = \{(U, \phi)\}$$ such that $$\phi:U\rightarrow\mathbb{R}^{n}$$ is a homeomorphism onto $$\phi(U)$$ and $$\phi(U)$$ is an open subset of $$\mathbb{R}^{n}$$ (where $$n$$ is the same for all charts here). The dimension of a topological manifold is the number $$n$$ that appears in $$\phi:U\rightarrow\mathbb{R}^{n}$$ for charts $$(U, \phi)\in\mathcal{A}$$.

A smooth manifold is a topological manifold but with a restricted atlas $$\mathcal{A}'\subseteq\mathcal{A}$$ such that any two chart maps $$\phi, \psi$$ from $$\mathcal{A}'$$ are "smoothly compatible" (i.e. $$\phi\circ\psi^{-1}$$ and $$\psi\circ\phi^{-1}$$ are infinitely differentiable functions). The dimension of a smooth manifold is the number $$m$$ that appears in $$\phi:U\rightarrow\mathbb{R}^{m}$$ for charts $$(U, \phi)\in\mathcal{A}'$$.

Since $$\mathcal{A}'$$ is simply a subset of $$\mathcal{A}$$, the number $$m$$ appearing in $$\mathcal{A}'$$ is the same as the number that appears in $$\mathcal{A}$$. Therefore, $$n = m$$.

Edit: As pointed out, the dimension of a manifold is a purely topological property. My definition for the dimension of a smooth manifold ends up being a topological property, and therefore it is redundant to define.

To answer the questions, #1 is yes. For #2, the answer is yes.

For #3, the answer to the first bullet is no. Consider the following counterexample.

Counterexample. Let $$M = \mathbb{R}$$ with the smooth structure given by the global chart $$(\mathbb{R}, \phi)$$ where $$\phi(t) = t$$. Then define the map $$F:M\rightarrow M$$ by $$F(r) = r^{3}$$. In coordinates this is $$\phi\circ F\circ\phi^{-1}(x^{1}) = (x^{1})^{3}.$$ One can check that $$F$$ is smooth homeomorphism (however, it is not a diffeomorphism because $$F^{-1}$$ has an infinite derivative at $$x^{1} = 0$$). For any $$r\in\mathbb{R}$$, the tangent space $$T_{r}M$$ is spanned by the basis vector $$\partial_{1}|_{r}$$. Then $$F_{*}(\partial_{1}|_{r}) = \frac{\partial F^{1}(r)}{\partial x^{1}}\, \partial_{1}|_{F(r)} = 3r^{2}\, \partial_{1}|_{r^{3}}.$$ At $$r=0$$, we see $$F_{*}(\partial_{1}|_{r}) = 0$$, so $$F_{*}$$ sends a $$1$$-dimensional vector space $$T_{0}M$$ to a zero dimensional subspace of $$T_{0}M$$, so it is not a vector space isomorphism.

For the second bullet point of #3, the answer is yes assuming the sets are nonempty.

Theorem. If a nonempty open set $$U\subseteq\mathbb{R}^{n}$$ is homeomorphic to a nonempty open set $$V\subseteq\mathbb{R}^{m}$$, then $$m=n$$.

Surprisingly, this theorem requires some heavy tools of topology. I don't know much of algebraic topology, so I can't be much help. However, I've found that this, this and this to be at least relevant.

• For the last point: I think the easiest way of showing that is assuming $U$ is a ball, picking an injective map $\phi:U\to V$ and looking at $\phi: U\setminus \{x\}\to V\setminus \{\phi(x)\}$ for some $x\in U$ and then looking at the induced map in homology. There is some technical detail in showing that $V$ must be homeomorphic to the $m$-dimensional ball, or at least that $V$ without a point has homology in degree $m$, which $U\setminus \{x\}$ does not have unless $m=n$. – WoolierThanThou Jul 20 at 10:58
• ...I guess what I'm sketching needs $n=1$ to be treated separately, but that should be pretty standard by, again, removing a point, and just considering connectivity. – WoolierThanThou Jul 20 at 11:02
• Thanks SpiralRain! For 3.2, these are open subsets of $\mathbb R^n$ and thus smooth manifolds? What do you mean? – Selene Auckland Jul 20 at 11:58
• @SeleneAuckland Because $\varnothing\subseteq\mathbb{R}^{2}$ is homeomorphic to $\varnothing\subseteq\mathbb{R}^{3}$ and in both situations $\varnothing$ is an open set. lol In other words, the empty sets violate the theorem. See this. – SpiralRain Jul 21 at 6:08
• @SeleneAuckland This is bit of a word game, but the answer is that they are literally the same set: they're both $\{ \}$. The empty function $h:\varnothing\rightarrow\varnothing$ which does nothing is a homeomorphism. This link with the answer should explain it a bit. – SpiralRain Jul 21 at 6:24

Not a complete answer, but a rephrasing.

It seems to me that this question can be reduced a good deal, to a far simpler question:

Suppose that $$M$$ is a smooth manifold of (smooth) dimension $$n$$, and a topological manifold of (topological) dimension $$k$$ with the property that the identity $$i : M \to M : x \mapsto x$$ is a homeomorphism from the topological manifold to the smooth manifold. Must $$n$$ and $$k$$ be equal?

For if so, then the invariance of topological dimension of a topological manifold under homeomorphism then proves the invariance of smooth dimension (as noted by others).