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I was going thorough the definition of a manifold and needless to say it wasn't something that I could digest at one go. Then I saw the following Quora link and Qiaochu's illustrative answer.

It was great to see the motivation behind the concept of manifold. Then I looked at the definition of manifold that I have at my disposal which is the following:

A topological space $M$ is an $n$-dimensional real manifold if there is a family of subsets $U_\alpha$, $\alpha \in A$, of $\mathbb{R}^n$ and a quotient map $f \colon \coprod_\alpha U_\alpha \to M$ such that $f|_{U_\alpha}$ is a homeomorphism onto the image for all $\alpha$.

I understand that we are trying to conceptualize about a bigger space which when looked at a very small region looks like something else (a euclidean space) thereby giving a possibly incorrect bigger picture about the shape. Now what was the reason behind introducing disjoint union in this definition. Can someone help me to get in terms with this idea? I went through this related question and Samuel's brilliant answer to it.

Leaving aside the points about Haussdorff and second countable spaces I could draw that homeomorphism is the concept that we use to convey the similarity between two spaces. I can loosely convince myself that the existence of homeomorphism between two spaces means the similarity in the pattern of open sets in the two spaces. (I might not be expressing what I feel about it.) But still then can anyone make it a bit more elaborate as to why we use homeomorphism here? If we divided this bigger surface, let's say earth into smaller circles, then I can see that we wouldn't have gotten the local shape same everywhere, somewhere it would have been a circle and some other points it would have been an area in between four circles or maybe something else. But are these local shapes homeomorphic to $\mathbb{R}^2$? What are other shapes which wouldn't have been homeomorphic to $\mathbb{R}^2$?

I guess "What happens when two spaces are homeomorphic?" would be a good way to start the discussion.

Community
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jedna dve
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  • Very interesting question! – Pedro May 21 '16 at 12:53
  • I agree with one of the answer below. The definition $f: \coprod_\alpha U_\alpha \to M$ is equivalent but less transparent of the one with the collection of maps $f_\alpha$ with the associated transition maps $ f_\beta\circ f_\alpha^{-1}$ (see https://en.wikipedia.org/wiki/Differentiable_manifold, paragraph "Atlas"). The central idea of manifold is double: local homeomorphisms with open subsets of $\mathbb R^n$ plus a way of gluing up those pieces. – guestDiego May 21 '16 at 14:02
  • @guestDiego : Could you elaborate a little bit. – jedna dve May 21 '16 at 17:29
  • Briefly, what I meant is that the collection of homeomorphisms $f_\alpha$ specify the local structure of the manifolds (essentially the dimension of the manifolds). The transition maps $f_\beta\circ f_\alpha^{-1}$ defined over the overlaps $U_\alpha\cap U_\beta$ of the covering specify the global structure of the manifold: so you can construct a sphere or a torus out of four pieces homeomorphic to a square of $R^2$. What changes are the transition maps. Observe also that, putting stricter restrictions on the transition maps, like smoothness, allows to define smooth manifolds as well – guestDiego May 22 '16 at 14:00

2 Answers2

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"Being homeomorphic" is a very strong condition for spaces - it means that we have a bijection between the underlying sets of the spaces that is continuous, and whose inverse is continuous. In particular, you also get a bijection between the open sets of the two spaces.

We can therefore consider two spaces that are homeomorphic to be "essentially the same" for the purposes of topology.

So what goes wrong with your "Earth" analogy? I suspect when you say "circle" you mean "disc" (a circle is e.g. $\{x \in \mathbb R^2 \: | \: |x| = 1\}$ while a(n open) disc is e.g. $\{x \in \mathbb R^2 \: | \: |x| < 1\}$). Circles aren't open in the normal topology on a sphere, so we definitely want to be thinking about discs.

You say "some other points it would have been an area in between four [disks] or maybe something else" - this is true, but when dealing with a manifold we're always careful to make sure that the open disks we choose completely cover the manifold. This means that every point $p$ is contained in some $U_\alpha$.

I think part of the confusion is that the definition you've given isn't (to my mind) the most natural. In my opinion the nicest way to approach it is as follows: a manifold is a topological space $M$, and a collection of open sets $\{U_\alpha : \alpha \in A\}$ that cover $M$ (so given $p \in M$ we can find some $\alpha$ with $p \in U_\alpha$), and a corresponding collection $\{V_\alpha : \alpha \in A\}$ of open subsets of $\mathbb R^n$, and a collection of homeomorphisms $\phi_\alpha: U_\alpha \to V_\alpha$. (There are also the hypotheses about Hausdorff, second countable etc., but these aren't key for understanding the idea of a manifold.)

This definition is equivalent to the one you gave, but it makes it clearer that what we're caring about is that at every point $p$, the manifold $M$ locally looks like a subset of $\mathbb R^n$.

I'm not sure if that completely answers your questions, feel free to comment if there was something more specific that was confusing you.

Josh Hunt
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  • Thanks a lot for your answer . – jedna dve May 21 '16 at 17:28
  • For each homeomorphism we again become concerned about the subsets of each $\ U_\alpha $ ,right ? – jedna dve May 21 '16 at 17:42
  • Yes, exactly - we have a homeomorphism from $U_\alpha$ to $V_\alpha$, so there is a bijection between the open subsets of $U_\alpha$ and the open subsets of $V_\alpha$ (which are nicely-behaved, because $V_\alpha \subseteq \mathbb R^n$). – Josh Hunt May 21 '16 at 22:40
  • I might be missing something , but would like to ask what if in the definition we had chosen a set of disjoint subsets of M , which cover it ? – jedna dve Jun 19 '16 at 16:32
  • That would only describe a very small portion of manifolds! Namely, the ones that are a disjoint union of open disks. Because you're picking a cover of open sets (not any old sets) the only connected manifold you could have under the new definition is the open disk itself. We definitely want to include more objects than this in our description of a manifold. – Josh Hunt Jun 19 '16 at 16:35
  • :Well , connectedness is still to be done by me . I guess I might not have expressed my question properly . I meant , we can always find a set of disjoint subsets which cover M , so why not use that in the definition . Is it because ,we might not find sets of homeomorphism under quotient topology that relate these disjoint subsets to subsets of $\ R ^n $ ? – jedna dve Jun 19 '16 at 17:19
  • The problem is still that the sets in the definition of a manifold need to be open sets, not any old sets. A topological space is defined to be connected iff there is no cover of it by disjoint open subsets, and so we can't find a cover by disjoint open subsets in a lot of cases that we care about. – Josh Hunt Jun 19 '16 at 17:30
  • So,we keep connectedness while defining a manifold. On a different note , did the guy who had defined manifold knew of the concept of "connectedness" or these concept developed together ? – jedna dve Jun 19 '16 at 19:09
  • Not quite - there are manifolds that are not connected. The point is that there are things that are connected that we would like to think of as manifolds, but wouldn't be allowed under your definition (for example the sphere $S^2$). I'm afraid I don't know the history of the definitions of a manifold, sorry. – Josh Hunt Jun 19 '16 at 19:12
  • Anyways ..thanks a lot . But could you clarify a bit more about using disjoint union but not taking the partition of set into disjoint subsets and defining homeomorphism from each such subset to a corresponding subset of $\ R^n$ ? – jedna dve Jun 22 '16 at 18:29
  • I'm sorry, I don't quite understand your question. Perhaps the easiest would be to start a new question here on mathexchange, explain clearly what it is you're confused about, and post a link to it here. – Josh Hunt Jun 22 '16 at 19:10
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The definition you've given is using a disjoint union as a sort of shorthand for a collection of maps. After all, one map $f: \coprod_\alpha U_\alpha \to M$ is equivalent to a collection of maps $\{f_\alpha: U_\alpha \to M\}$ where $f_\alpha = f|_{U_\alpha}$. The $f_\alpha$ are the charts for the manifold which identify an open subset of $M$ with an open subset of $\Bbb R^n$.

The definition of manifold which I'm more familiar with is the following: A topological space $M$ is an $n$-dimensional manifold if there exist open subsets $U_\alpha \subseteq \Bbb R^n$, $V_\alpha \subseteq M$, and homeomorphisms $f_\alpha: U_\alpha \to V_\alpha$ such that $\bigcup_\alpha V_\alpha = M$ (i.e. every point in $M$ is in some chart $V_\alpha$). We also assume that $M$ is second countable and Hausdorff. Personally, I find this definition to be a bit more transparent. It's easier to visualize what this definition is saying.

Now, for your question about what a homeomorphism really means. Intuitively, two spaces are homeomorphic if they are completely equivalent for the purposes of topology. Any topological construction you make, or topological property you prove about one space will hold for a homeomorphic space. They really are exacly the same.

More technically, a homeomorphism $f: X \to Y$ is firstly a bijection of sets, so it shows that if you start with $X$ and just relabel the points, then you get $Y$, as a set. More than that though, $f$ also induces a bijection on the sets of open subsets. Let $\tau_X$ be the topology on $X$, i.e. the set of all open subsets of $X$. Likewise for $\tau_Y$. Then we have a bijection $f: \tau_X \to \tau_Y$ given by $U \mapsto f(U)$. Therefore, the homeomorphism $f$ shows that if you start with $X$ and relabel its open subsets, you end up with the open subsets of $Y$. So if all you care about is the collection of open subsets of a space (i.e. you're doing topology), then homeomorphic spaces only differ by how you labeled the points and open subsets (which obviously doesn't make any tangible difference).

Alex G.
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  • :Thanks a lot for your answer . – jedna dve May 21 '16 at 17:28
  • G : I guess the homeomorphism between two spaces will again depend upon the topology you are choosing . Like , if I ask whether a circle is homeomorphic to a square or whether a square is homeomorphic to a triangle. Then , if you take each and every point to be an open set ,then possesing equal number of points will make them homeomorphic . Isn't it so ? And in this way I guess any topological space can be called a manifold .( I am loosely saying this last statement,I guess for finite spaces it won't hold). – jedna dve May 21 '16 at 17:40
  • @jednadve No, not every topological space is a manifold. A topological space is a set with a fixed topology. You can't just choose a different topology and then call it a manifold. That would make it a different topological space. And actually, finite spaces $X$ with the discrete topology are manifolds. They are $0$-dimensional manifolds, and their charts are the functions $\Bbb R^0 \to {x}$, where $x$ is any point in $X$. – Alex G. May 21 '16 at 21:24
  • Thank you . I would like to digress slightly for a possibly related question . When people say a square can be transformed into a rectangle or a cylinder into a rectangle , do they mean the existence of homeomorphism between them ? I guess in that case we have no decided topology for them , so if we choose each and every point as open set , then there will be a homeomorphism and any shape can be transformed into any other shape . Isn't it so ? This shape transformation thing is a bit unclear to me . – jedna dve May 22 '16 at 05:17
  • In such cases , what rules decide which shape can be transformed into which ones ? – jedna dve May 22 '16 at 05:18