Timeline for Abstract connectedness
Current License: CC BY-SA 3.0
33 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 11, 2015 at 17:13 | answer | added | Joseph Van Name | timeline score: 3 | |
| Mar 9, 2015 at 19:00 | vote | accept | Mike Shulman | ||
| Mar 8, 2015 at 5:30 | comment | added | Mike Shulman | @QiaochuYuan - So are you suggesting that a "connectedness space" would be a set together with an equivalence relation on all of its subsets? There would then presumably have to be some axioms relating those equivalence relations to each other... | |
| Mar 7, 2015 at 17:35 | answer | added | Todd Trimble | timeline score: 12 | |
| Mar 7, 2015 at 5:33 | comment | added | Qiaochu Yuan | @Mike: of course any subspace also has a "same connected component" equivalence relation. Can you be more specific about what sorts of things you'd like to be able to say / prove using "abstract connectedness"? | |
| Mar 7, 2015 at 5:08 | comment | added | Mike Shulman | @QiaochuYuan I don't agree that "anything one might want to say about connectedness" is contained in that equivalence relation. See Eric's comment above: there are connected subsets that are not components, and subsets of connected components that are not connected; and just knowing which sets are the connected components doesn't tell you anything about those sets. | |
| Mar 6, 2015 at 23:29 | comment | added | Qiaochu Yuan | @Mike: maybe I don't understand the question. On a topological space, there's an equivalence relation called "being in the same connected component." Anything one might want to say about connectedness is contained in the data of this equivalence relation. (Again, I don't understand what your precisification has to do with the question as it is stated in the first paragraph, so I don't think I understand what you're actually looking for.) | |
| Mar 6, 2015 at 22:33 | comment | added | sure | @MikeShulman: nothing interesting, I was just reinventing what is apparently already well known. I was proposing to take groupoids as the standard definition of spaces with many connected components, and functors between them as "connected maps". | |
| Mar 6, 2015 at 22:06 | comment | added | Tobias Fritz | @MikeShulman: yes, I strongly agree about the minimality of the separoid axioms! It feels like they're missing something important... | |
| Mar 6, 2015 at 21:56 | comment | added | Mike Shulman | @TobiasFritz that's different from the specific question I asked in my second paragraph, but I might still consider it an answer to my third paragraph. I guess in a separoid you can define a subset to be "connected" if it is not the union of two nonempty separated subsets, and any morphism of separoids will preserve those? The axioms of a separoid do seem extremely minimal, though, so they probably aren't very close to the image of Top. | |
| Mar 6, 2015 at 21:54 | comment | added | Mike Shulman | @Hurkyl What does it generalize, and to what? | |
| Mar 6, 2015 at 21:53 | comment | added | user13113 | @Mike: It generalizes easily. The question seemed a bit complicated and it wasn't obvious to me what your actual goal was, so I wanted to do a quick check to see if you've already thought about the simple thing. | |
| Mar 6, 2015 at 21:51 | comment | added | Mike Shulman | @Hurkyl Sorry, what I meant was, what was your point in rephrasing the definition that way? | |
| Mar 6, 2015 at 21:42 | comment | added | user13113 | @Mike: In Top, the definition I gave is the standard topological definition; (the images of) $Y$ and $Z$ are disjoint clopen subsets of $Y \amalg Z$, and if $U$ and $V$ are disjoint clopen subsets of $X$ with $U \cup V = X$, then $X \cong U \amalg V$. | |
| Mar 6, 2015 at 21:36 | comment | added | Tobias Fritz | The notion of "separoid" may be related: en.wikipedia.org/wiki/Separoid "Given a topological space, we can define a separoid saying that two subsets are separated if there exist two disjoint open sets which contains them (one for each of them)." But possibly it captures something slightly different from what you're looking for; I haven't thought about it. | |
| Mar 6, 2015 at 21:22 | answer | added | Włodzimierz Holsztyński | timeline score: 1 | |
| Mar 6, 2015 at 21:06 | comment | added | Mike Shulman | @Hurkyl I'm using the standard topological definition of "connected". | |
| Mar 6, 2015 at 21:05 | comment | added | Mike Shulman | @sure, I'm having trouble putting together your sequence of comments into a coherent picture. Maybe you could write up your proposal coherently and post it as an answer? | |
| Mar 6, 2015 at 21:05 | comment | added | Mike Shulman | @QiaochuYuan can you explain in what way "equivalence relations" answers the question? | |
| Mar 6, 2015 at 21:04 | comment | added | Mike Shulman | @ToddTrimble - I generally do not consider the empty set to be connected, but I suppose I'm willing to be flexible if necessary. | |
| Mar 6, 2015 at 19:47 | comment | added | user13113 | Possibly silly question: have you considered "$X$ is connected" to mean that if $X \cong Y \amalg Z$ then $Y$ or $Z$ is initial? | |
| Mar 6, 2015 at 18:08 | comment | added | Qiaochu Yuan | Without looking at the way you've made the question precise, my immediate reaction to the question as it's first phrased is "equivalence relations." Is there more to say than this? | |
| Mar 6, 2015 at 16:22 | answer | added | Buschi Sergio | timeline score: 0 | |
| Mar 6, 2015 at 13:06 | comment | added | sure | You're right. I don't know if there exists a characterization of $[0,r]$ as a poset category, but I guess you could even define pathes of length $r$ in $G$ as functors from $[0,r]$ to $G$. Moreover, if $0$ is the trivial category, and $\{0,1\}$ seen as a discrete category, what would be the pushout of $[0,1] \leftarrow \{0,1\} \rightarrow {0}$ in Cat? Do we obtain something that "looks like" $S^1$ but as a category? If yes, could we also define some cone and suspension functor, in order to define higher dimensional pathes? | |
| Mar 6, 2015 at 11:42 | comment | added | Todd Trimble | Do you want the empty set to be connected? :-) [It may seem a trivial point, but ruling it out might aid in getting a smooth axiomatics.] | |
| Mar 6, 2015 at 9:49 | comment | added | sure | In particular, a path in such a space $G$ could be defined as a functor from any "ordinal" categories to $G$.That is, a path is nothing else than a diagram of type $* \rightarrow *' \rightarrow *'' \rightarrow .... \rightarrow *^x$ in $G$. Clearly, for any path $f$ there exists an opposite path $-f$ (the opposite one), there is associativity by concatenation of diagrams, and identity path, so we can form a category whose objects are the objects of $G$ and arrows the paths defined previously. Such category could be called the foundamental groupoid of connected groupoids? :< | |
| Mar 6, 2015 at 9:36 | comment | added | sure | With such definition, because a functor preserves isomorphisms, it is automatic that any functor from the (disjoint) sum of connected groupoids to another (disjoint) sum of connected groupoids sends connected components to connected components (that is, connected groupoids to connected groupoids). | |
| Mar 6, 2015 at 9:26 | comment | added | sure | Couldn't we take as connected space, any connected groupoid? That is, any groupoid $G$ such that for all $x$,$y$ in $Obj(G)$, $Arr(x,y)$ is non empty? A "connected" map will then by any functor from connected groupoids. (One could say that this captures the notion of path-connectedness, but is the distinction really relevant? :<) | |
| Mar 6, 2015 at 8:36 | answer | added | Włodzimierz Holsztyński | timeline score: 8 | |
| Mar 6, 2015 at 7:47 | comment | added | Eric Wofsey | @JochenWengenroth: The collection of all connected sets contains far more information than just the partition into connected components, because not every subset of a component is still connected. | |
| Mar 6, 2015 at 7:39 | comment | added | Jochen Wengenroth | Two connected components are either equal or disjoint, hence they form a partition of the underlying set. One could thus consider the full subcategory of pairs $(X,C_X)$ where $C_X$ is a partition of $X$. Given a partition $C_X$ ox $X$ the system of all unions $\bigcup M$ with $M\subseteq C_X$ is then a topology so that the elements of $C_X$ are the connected components. | |
| Mar 6, 2015 at 7:30 | comment | added | Eric Wofsey | For an example of a non-obvious property that connected subsets of a topological space satisfy, see the theorem mentioned in this answer. | |
| Mar 6, 2015 at 5:35 | history | asked | Mike Shulman | CC BY-SA 3.0 |