I'm studying abstract topological manifolds, and a way to understand such a structure is as a topological space that locally "resembles" Euclidean space. More rigorously, we want that for any point $p$ in our manifold, there is a neighborhood which can be mapped homeomorphically to $\mathbb{R}^{k}$ (in which case we'd have an $k$-dimensional manifold). We can make modifications for manifolds with boundary by introducing the half-space, but I don't need that addition for the purposes of my question.

My question is as follows: how do homeomorphic sets look in comparison to each other? How does the fact that neighborhoods in our manifold are homeomorphic to subsets of $\mathbb{R}^{k}$ make them look similar?

  • 1
    $\begingroup$ What do you mean by "look similar", or "look in comparison to each other"? Topologically speaking, these neighborhoods are all the "same", in that they have the same topological properties, namely those of $\mathbb{R}^k$. $\endgroup$ – Ben Steffan Mar 21 '20 at 17:32
  • $\begingroup$ @BenSteffan yes, but they can be compared with suitable definitions $\endgroup$ – user757601 Mar 21 '20 at 17:46
  • $\begingroup$ They can "look" quite different. In $\Bbb R$ the three subspaces $\Bbb Q,\,\Bbb Q\cap [1,\infty),\,\Bbb Q\cap [0,1]$ are homeomorphic to each other. $\endgroup$ – DanielWainfleet Mar 22 '20 at 7:14

The standard interpretation of homeomorphic subsets of $\mathbb{R}^n$ is one than can be deformed into one another without tearing them.

In $\mathbb{R}^2$ much more is true: the Riemann mapping theorem gives that any two non-empty, open simply connected subsets are homeomorphic. However, this fails for higher dimensions.

A homeomorphism is a map, which is continuous with continuous inverse. Hence if we want to interpret a continuous map as a map which "does not eliminate any nearness relations, but make create some", this gives the intuition that if two spaces are homeomorphic, then two points are close to one another if and only if they are close in the other.

An illustrative example is $[0,2\pi) \subseteq \mathbb{R}^2$ and $\mathbb{S}^1 \subseteq \mathbb{R}^2$ and the map $\phi:= e^{it}: [0,2\pi) \rightarrow \mathbb{S}^1$. Then we have that $\phi$ is continuous, but does not have a continuous inverse. Note that it preserves all the "nearness relations", but it also creates one: $\phi(0)$ is near to $\phi(2\pi)$. If there were to be an inverse, then it would need to map $\phi(0)$ to $0$ and $\phi(2 \pi)$ i.e. take two points which are "near" and map them to some which are not near anymore i.e. eliminate a "nearness relation". Hence two spaces are homeomorphic if there exists a correspondence between their points s.t. all the "nearness relations" are preserved and no new ones are created.

Another approach is the following: A map is continuous if and only if preimages of open sets are open. Hence if two spaces are homeomorphic there is a one to one correspondence of open sets. How does tie into the intuition before? In a metric space, the open balls generate the topology, hence for two spaces to be homeomorphic, it is actually sufficient to require that there is a one to one correspondence of the open balls. However, now note that two points are in the same open ball if their distance is smaller than the radius of the open ball. Hence if all open balls coincide, then all distance relationships coincide (note that that does not mean that all the distances coincide though. One can only conclude that for two points $x,y$ which are "close" we also have that $\phi(x), \phi(y)$ are "close") an illustration of this is the fact that $(0,1)$ and $\mathbb{R}$ are homeomorphic. Though, whatever homeomorphism we choose, there are some points $x,y \in (0,1)$ s.t. $\phi(x)$ and $\phi(y)$ have distance greater than $2$ despite $x$ and $y$ being distance less than $1$ apart.

Finally, the constraint that a manifold that it is patched up by such open subsets of $\mathbb{R}^n$.

  • 1
    $\begingroup$ I became less fond of the "deformed" terminology after realizing that the circle and the trefoil knot are homeomorphic subsets of $\mathbb{R}^3$. $\endgroup$ – Nate Eldredge Mar 21 '20 at 17:51
  • $\begingroup$ @NateEldredge: Yes, they are. And it makes sense: If you could not see the space around them, how would you distinguish a circle from a trefoil knot? Note also that if you embed the $\mathbb R^3$ into $\mathbb R^4$ (which certainly doesn't change the topology of subsets of $\mathbb R^3$), you can indeed continuously deform the circle into the trefoil knot. $\endgroup$ – celtschk Mar 21 '20 at 17:58
  • $\begingroup$ To be honest, I'm not that big of a fan of that interpretation either. But I suppose for simples subset it is sufficient and I think it especially useful when first learning about homeomorphisms. Even more so when talking about manifolds. $\endgroup$ – G. Chiusole Mar 21 '20 at 17:59
  • $\begingroup$ The word "deform" usually leads to the impression that there is an ambient space in which the deformation happens. This often causes confusions to beginners and non-mathematicians. A manifold can exist without being embedded into an ambient space. $\endgroup$ – WhatsUp Mar 21 '20 at 18:53
  • $\begingroup$ Again, I'm not quite happy with it either. But I don't think that it leading to the impression that there is an ambient space is all that bad. After all, it is always possible to embedd it into one. Also, I don't think that it causes confusion for beginners - especially considering that a considerable abount of people learning about manifolds do so before a course in general topology (physicists, engineers, ...). Of course, the best explanation is: "open sets correspond" or "all topological properties coincide", but thats hardly intuitive. $\endgroup$ – G. Chiusole Mar 21 '20 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.