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Given a set $M$, we can construct a topology $\tau$ to this set, which is nothing but a collection of open sets, making it a topological space $(M,\tau)$. There are many ways to do so.

Continuity is defined based on the concept of open sets, i.e., a function $f:(M,\tau) \rightarrow (N,\tau')$ from one topological space to another is said to be continuous if and only if the preimage of each open set of $N$ is open. There is here something which is pretty awkward to me: the notion of continuity is not a property of the function itself, but depends on the topologies you choose for $M$ and $N$.

Since topologies change the continuous character of functions, this makes homeomorphisms from two different sets/manifolds not to be a intrinsic property between the two sets/manifolds, i.e., it depends on the topology you choose.

For instance, pick $\mathbb{R}^{n}$ with euclidian metric $|x| = \sqrt{\sum\limits_{i=1}^{n} x_{i}^{2}}$. We can construct a topology by choosing a basis with the n-balls $B(p;r) = \{x \in \mathbb{R}^{n}; \, |x-p|<r\}$. Let's call this topology $\tau$. Now, let $\tau_{0}$ be the trivial topology for $\mathbb{R}^{n}$, i.e., $\tau_{0} = \{\mathbb{R}^{n}, \varnothing \}$. Let $f:(\mathbb{R}^{n},\tau_{0}) \rightarrow (\mathbb{R}^{n}, \tau)$ be given by $f(x)=x, \forall x \in \mathbb{R}^{n}$. One can see that $f$ is not a homeomorphism since it is not continuous.

Nonetheless, we know that surfaces in $\mathbb{R}^{3}$ can be topologicaly identified in terms of its genus. I.e., We have a list of surfaces (sphere, torus, 2-torus, ..., n-torus,...) that make any other surface homeomorphic to one of them according to its genus. And genus is something intrinsic of the surface. Therefore the topology is intrinsic to the surface. Thus we have here a contradiction with the construction made above.

So, my question is: Is topology a intrinsic property of manifolds or is it an arbitrary construction with respect to nothing but the definition of open set?

And, if it is indeed intrinsic, what is the meaning of saying a topology when there is only one topology for a manifold?

Thank you!!

Eric Wofsey
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Jean Weigel
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    Manifolds are not sets. Manifolds have way more structure than that. In particular, manifolds are themselves topological spaces. You can give their underlying set a different topology, sure. But then it's no longer the same manifold, if it is still a manifold at all. – Vercassivelaunos Sep 02 '21 at 19:54
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    Usually, we define a topology on something intrinsic. So the topology of a circle is definitely some "about" our intuition of a circle. There are lots of topologies on that same set of points, but none of the topologies capture what we understand about a circle. So yes, a topology does capture something intrinsic. It is just not intrinsic to the set. But it shows there are intrinsic things about geometry and spaces which are more than simply a set of points in the space. – Thomas Andrews Sep 02 '21 at 20:09
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    This is a bit like saying a ring isn't intrinsic to a set. Of course it isn't. A ring is a set with certain operations, addition and multiplication, and certain elements, with an additive and multiplicative identity and a few more conditions. They are also not "intrinsic" the the set of elements of, say, $\mathbb Z.$ – Thomas Andrews Sep 02 '21 at 20:12
  • @Vercassivelaunos So when we say, for example, that $\mathbb{S}^{2}$ is a surface in $\mathbb{R}^{3}$ ( which is a topological manifold with more properties) what topology we implicitly give it? The subspace topology? Because seems to me, when people talk about $\mathbb{S}^{2}$ it is just the set ${ x \in \mathbb{R}^{3}; , |x|=1}$ with an atlas that has certain properties, although I know the concept of topological manifolds is given to second countable hausdorff spaces that are locally euclidean. – Jean Weigel Sep 02 '21 at 20:48
  • @Vercassivelaunos furthermore, for a topological space $(M,\tau)$ to be locally euclidean, there must be homeomorphisms from each neighborhood of $(M,\tau)$ to $\mathbb{R}^{n}$, isn't we assuming some topology for $\mathbb{R}^{n}$ here? – Jean Weigel Sep 02 '21 at 20:52
  • @ThomasAndrews Ok, different topologies on the same set of points make different topological spaces, but the circle itself isn't just the set ${x \in \mathbb{R}^{2}; , |x|=1}$? "but none of the topologies capture what we understand about a circle. So yes, a topology does capture something intrinsic." - I do not understand. – Jean Weigel Sep 02 '21 at 21:00
  • @ThomasAndrews For example, if you have two different metrics $g, g'$ on the same (differentiable) manifold $M$, you have two different Riemannian Manifolds $(M,g)$ and $(M,g')$. Their geometry is completely distinct. However, when you're dealing with topology, it seems to me it's not the same thing. Because holes, compactness, homeomorphisms seems to me, intuitively, they are properties of the set underlying the topological space, not something that comes from choosing a topology. – Jean Weigel Sep 02 '21 at 21:06
  • They aren’t differentiable manifolds until you have a topology, so it isn’t two different metrics on the manifold, it’s two different metrics which make a set of points a manifold. @JeanWeigel – Thomas Andrews Sep 02 '21 at 21:08
  • But sets don’t have holes, either. The question is, how do we clarify spaces in terms of holes? You can’t do it with points alone, you need some additional information. That information turns out to be topology. It’s not obvious that this is the case, but it is true. – Thomas Andrews Sep 02 '21 at 21:08
  • "That information turns out to be topology" - but this doesn't mean we can make two topological spaces $(M,\tau_{1})$ and $(M,\tau_{2})$ such that one has holes and the other does not have holes? – Jean Weigel Sep 02 '21 at 21:14
  • The assumed topology of $\mathbb R^n$ is the usual Euclidean topology. When talking about $\mathbb S^2$ as a surface in $\mathbb R^3$, we then use the subspace topology. And regarding your last question directed at Thomas: one certainly can. We can make the set $\mathbb R$ look like a circle. Just choose your favorite bijection between the two, and define that a set in $\mathbb R$ is open iff its image under the bijection is open. – Vercassivelaunos Sep 02 '21 at 22:37

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Topology arose when mathematicians realized continuity, which they initially defined in terms of metrics, was not really about a metric, but about the open sets. Two metrics can give the same notion of continuity, and they do so when they give the same notion of open sets.

So mathematicians defined topology. Originally, they included a lot of extra conditions. Early definitions, I know, were basically Hausdorff spaces, possibly with some more conditions.

But then people found more things that were like topologies, but didn’t satisfy some of the axioms, and couldn’t be seen in terms of metrics. And it was useful to discuss functions there that were “continuous.” So they started loosening the definition of topology.

All of your intuition about spaces has an intrinsic metric associated with it. A set of points without that intuition is really only determined by its cardinality - all sets of the cardinality of the continuum are essentially the same when we treat them as sets.

What mathematicians realized was that there is no need for a metric when we are studying things that are not strictly tied to lengths. The simplest example of this is that the real line’s continuous functions are entirely determined by the $<$ relation on the real line, without any metric.

Differentiable manifolds have even more structure, to capture our intuition about them, to aid in studying of them. The point set topology only begins to cover our intuition related to them, and the things we want to study about them.

When I think of a geometric torus, I think of the two circles involved as being different. But if I think of it as $X=\mathbb R^2/\mathbb Z^2,$ it is obvious that they aren’t. We can define a distance on $X,$ of course, but we don’t need it to get a topological torus, if we already have a topology on $\mathbb R^2,$ and the group properties of addition there. That gives us a powerful tool for looking at such spaces. We now see that the space as “flat” in some way - that every point is roughly the same. The “holes” we envision in $3$-dimensions are something else when looked at this way - it turns out, it is a property of the topology, not the $3$d geometry of a representation of the object.

An example of a quotient topology that doesn’t have an easy-to-define metric.

Consider $X=(-1,1)$ with the usual metric. Say for $x,y\in X$ that $x\sim y$ if and only if $\tan(\pi x/2)-\tan(\pi y/2)$ is an integer. Then $X/\sim$ does not have an “obvious” metric. You can only find the metric by considering that this is essentially the same as $\mathbb R/\mathbb Z=S^1.$ But we can define $X/\sim$ easily as a quotient topology without finding the related distance.

Mathematicians like generalizations. We started with metric spaces. Then topologies for a broader definition of continuity.

This article talks about a generalization to topology that is so technical, I can’t understand it, but it apparently lets mathematicians treat results in different fields as one result.

Thomas Andrews
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