[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.

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"]

  • $\begingroup$ @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** $\endgroup$ – TheQuantumMan Sep 13 '17 at 19:54
  • $\begingroup$ ** conditions, of course] $\endgroup$ – TheQuantumMan Sep 13 '17 at 19:54
  • $\begingroup$ @EricWofsey the link you posted in the comment actually had a helpful answer in that might shed some light $\endgroup$ – TheQuantumMan Sep 13 '17 at 19:56
  • $\begingroup$ 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: math.stackexchange.com/questions/761917/… $\endgroup$ – Eric Wofsey Sep 13 '17 at 19:57
  • $\begingroup$ 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. $\endgroup$ – Jack Lee Sep 14 '17 at 18:26

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 $\mathbb{R}^n$.)

  • $\begingroup$ 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) $\endgroup$ – TheQuantumMan Sep 13 '17 at 20:10
  • $\begingroup$ You need to have a topological manifold. I've added an extra paragraph on this. $\endgroup$ – Eric Wofsey Sep 13 '17 at 20:53
  • $\begingroup$ 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 ** $\endgroup$ – TheQuantumMan Sep 14 '17 at 22:14
  • $\begingroup$ **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? $\endgroup$ – TheQuantumMan Sep 14 '17 at 22:14
  • $\begingroup$ Your professor is just wrong, or you are understanding them wrong. $\endgroup$ – Eric Wofsey Sep 14 '17 at 22:17

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