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[Sorry if the following is not rigorous enough as I am an undergrad physicist with no background in pure mathematics. Please correct me if I stated something wrong or imprecisely]

In a course in Riemannian geometry, our professor gave us a definition of topological manifolds that (partly) had to do with the charts ${(U_a,\phi_a)}$ that covered the manifold and whose corresponding maps $\phi$ are homeomorphisms to $R^n$.

The professor then proceeded on to defining a differentiable manifold using the usual condition that the transition maps $\phi_a^{-1} \circ \phi_ {\beta}(W)$ with $U_{\alpha} \cap U_{\beta}=W\ne \emptyset$ need to be diffeomorphisms(homeomorphisms and $C^{\infty}$).

Now, in practice, to show that a topolgical manifold is smooth, we show the maps $\phi_{\alpha}$ are one-to-one, that the charts $(U_{\alpha},\phi_{\alpha})_{\alpha}$ cover the manifold and show that all $\phi_a^{-1} \circ \phi_ {\beta}(W)$ with $U_{\alpha} \cap U_{\beta}=W\ne \emptyset$ are diffeomorphisms.

But, since a smooth manifold is also a topological manifold, then why don't we also have to show that all the $\phi_{\alpha}$ are homeomorphisms since this is a necessary condition to have a topological manifold?
If we show this, and also show that the transition maps are $C^{\infty}$ then the transition maps are diffeomorphisms, as we want to show. But, this does not always guarantee that the maps $\phi_{\alpha}$ are homeomorphisms.

EDIT:
So, why should we just show that all the $\phi_{\alpha}$ are one-to-one? Why is it not needed to show that the coordinate maps $\phi_{\alpha}$ are onto and continuous with continuous inverse(since these along with the 1-1 condition tell us that they are homeomorphisms)?

What I said above is also true when we want to show that a topological space is a smooth manifold. So, the question is even more general.

[EDIT 2: I have edited the title due to a comment pointing out a mistake I have made. I switched from "topological manifold" to "topological space"]

TheQuantumMan
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  • @EricWofsey based on the answer, it seems that what I describe above does not have to do with showing that a topological manifold is a smooth manifold but it is more general. As I understand it(the answer) it's two ways of showing that a topological space is a topological manifold. Right? [even if it is right, I don't understand the reason for this "equality" of the two ways to prove it; i.e. via showing that the coordinate maps are homeomorphic or showing that the coordinate maps are bijective and the transition maps are homeomorphic [for smooth manifolds we also need differentiability** – TheQuantumMan Sep 13 '17 at 19:54
  • ** conditions, of course] – TheQuantumMan Sep 13 '17 at 19:54
  • @EricWofsey the link you posted in the comment actually had a helpful answer in that might shed some light – TheQuantumMan Sep 13 '17 at 19:56
  • I deleted it because I realized I had misread your question and the link I gave wasn't as closely related as I thought at first. But here it is in case it is still helpful: https://math.stackexchange.com/questions/761917/link-between-a-topological-space-and-a-manifold – Eric Wofsey Sep 13 '17 at 19:57
  • It's important to note that "how to show that a topological manifold is smooth" is not a meaningful question. A smooth manifold is a topological manifold with additional structure -- a smooth structure -- that is not determined by the topology. For more about this, see my answer to this question and Chapter 1 of my Introduction to Smooth Manifolds. – Jack Lee Sep 14 '17 at 18:26
  • @JackLee wow, thanks for the clarification! So, the answer to the question is that I first have to show that the set I have is a topological manifold and then prove that it has also got a smooth structure, right? If this is the case, I would first prove that the coordinate maps I chose are homeomorphisms. Then, to prove that my topological manifold has a smooth structure, I would prove that the maps are compatible, are $C^{\infty}$ and they cover the manifold. But, in practice, our professor told us that we just have to show that the coordinate maps are 1-1, compatible and** – TheQuantumMan Sep 14 '17 at 22:09
  • **cover the manifold. But, this is different from the first "proceedure" I described above, since in the second "procedure" we don't show that the coordinate maps are homeomorphisms(just 1-1). So, why do we not have to show that the coordinate maps are onto and continuous with continuous inverse? – TheQuantumMan Sep 14 '17 at 22:11
  • @TheQuantumMan: In my Smooth Manifolds, book, there's a lemma (Lemma 1.35, the Smooth Manifold Chart Lemma) that shows how to combine it into one step. You start with a set, together with a collection of mappings from subsets into R^n, and if they satisfy certain conditions they determine a topology and a smooth structure. Check it out. – Jack Lee Sep 14 '17 at 23:54
  • @JackLee I looked it up in the sample chapter you made available(thanks by the way; I already ordered your book on Riemannian geometry and I will be ordering your smooth manifolds book also!). So, indeed, one can skip showing that the coordinate maps are continuous and their inverse is continuous(it is implied by the compatibility property I suppose) and the "onto" part can be trivially satisfied with suitable choices of $U_{\alpha}$. Thank you(and please, do correct anything that I might have stated incorrectly). – TheQuantumMan Sep 15 '17 at 00:23

1 Answers1

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If you do not know that the $\phi_\alpha$ are continuous, then it does not suffice to just check that they are injective and the transition maps are diffeomorphisms. For instance, let $M$ be any topological manifold and let $\phi:M\to\mathbb{R}^n$ be any bijection (not necessarily even continuous!). Taking $\phi$ as our only chart on $M$, the only transition map is the identity $\mathbb{R}^n\to\mathbb{R}^n$ which is a diffeomrphism. But this certainly doesn't make $M$ a smooth manifold.

On the other hand, if you know that the $\phi_\alpha$ are continuous (but not necessarily homeomorphisms) it suffices to check that they are injective by invariance of domain, a hard theorem in topology. Specifically, invariance of domain says that if $U\subseteq\mathbb{R}^n$ is open and $f:U\to \mathbb{R}^n$ is a continuous injection, then $f$ is an open map.

Now in your case, suppose we have $\varphi_\alpha:U_\alpha\to\mathbb{R}^n$ which is injective and continuous. Since we are assuming we have a topological manifold (of dimension $n$), $U_\alpha$ is a union of open subsets that are homeomorphic to open subsets of $\mathbb{R}^n$. By invariance of domain, $\varphi_\alpha$ is an open map when restricted to each of these open subsets, and it follows that $\varphi_\alpha$ is an open map on all of $U_\alpha$. It is therefore a homeomorphism to its image.

It is crucial to this argument, though, that you already know you have a topological manifold. If you just have some arbitrary space, there are easy counterexamples. Indeed, similar to in the first paragraph, you can just take $M$ to be any topological space with a continuous bijection to $\mathbb{R}^n$ which is not a homeomorphism and use that as your only chart. (Such a space is easy to construct; for instance, let $M=\mathbb{R}^n$ and your bijection be the identity, and give $M$ a topology by adding some new open sets to the standard topology on $\mathbb{R}^n$.)

The Phenotype
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Eric Wofsey
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  • What if we don't assume that we have a topological manifold? I should have stated it in my question, but we also do what I described to show that a topological space is a smooth manifold(for example, we did this to show that $S^2$ is a smooth manifold) – TheQuantumMan Sep 13 '17 at 20:10
  • You need to have a topological manifold. I've added an extra paragraph on this. – Eric Wofsey Sep 13 '17 at 20:53
  • So, if I need to have a topological manifold, this means that I first have to show that the coordinate maps are homeomorphisms. And then proceed on to show that the maps are compatible and cover the manifold if we are to have a smooth manifold(i.e. topological manifold with smooth structure). But, in practice, our professor told us that we just have to show that the coordinate maps are 1-1, compatible and cover the manifold. But, this is different from the first "procedure" I described above, since in the second "procedure" we ** – TheQuantumMan Sep 14 '17 at 22:14
  • **don't show that the coordinate maps are homeomorphisms(just 1-1). So, why do we not have to show that the coordinate maps are onto and continuous with continuous inverse? – TheQuantumMan Sep 14 '17 at 22:14
  • Your professor is just wrong, or you are understanding them wrong. – Eric Wofsey Sep 14 '17 at 22:17
  • So, indeed, I must show that the coordinate maps are homemorphisms and not just 1-1. Thank you. – TheQuantumMan Sep 14 '17 at 22:36