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Assume that $X$ is a compact Hausdorff space and $A\subset X$ is a retract of $X$.

Is there a topological groupoid structure on the topological pair $(X,A)$ where, in the corresponding small category, $X$ and $A$ plays the role of morphisms and objects, respectively.

Edit 1: Is there a theory which investigate such type of problems for $A$ not necessarily a single point

In particular, is there a non single retract $A$ of $X=S^{3}$ or $X=S^{7}$ such that $(X,A)$ does not admit a topological groupoid structure.

In this question we do not require that the retracting map has any relation to the source and range maps

Edit 2: As another particular case, is there a groupid structure on $(G,G^{0})$ where $G=Gl(n,\mathbb{R})$ and $G^{0}=O(n)$? That is: are there range, source and composition maps which are continuous which makes $(G,G^{0})$ a groupoid.

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    $\begingroup$ For $A$ a point you are asking whether there is a topological group structure on any compact Hausdorff space, with given point as a unit. Obviously not. $\endgroup$ Commented Aug 15, 2016 at 13:33
  • $\begingroup$ @მამუკაჯიბლაძე thank you. What about if we add "Non singleton A"? $\endgroup$ Commented Aug 15, 2016 at 13:39
  • $\begingroup$ and X is connected $\endgroup$ Commented Aug 15, 2016 at 13:42
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    $\begingroup$ These sorts of dodges accomplish nothing: in any groupoid, there is a group $\hom(x, x)$ for each object $x$, imposing a heavy homogeneity condition which need not be satisfied in such generality. More importantly, if I am a researcher on this, the first question I ask is: what would the domain and codomain functions $X \to A$ be in terms of the data? I am supposing we are given a retraction $r: X \to A$ of the inclusion $i: A \hookrightarrow X$, but I can't seem to cook up a second function (unless the domain and codomain functions are actually the same). Then what would composition be?? $\endgroup$ Commented Aug 15, 2016 at 13:51
  • $\begingroup$ @ToddTrimble According to the last statment of your comment I would like to compare the situation with the topological group case: We have an space X then we ask is there a topological group structure on X. In this question we do not confirm any operation. No in the topological groupoid setting we have a pair we ask about a topological groupoid structure. So in our question we do not impose any r or s maps: We ask are there range source and composition map making the pair into a groupoid. I think I removed the misunderstanding, yes? $\endgroup$ Commented Aug 19, 2016 at 21:51

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Note: This answer had been sent to Ali in private email (the question had been closed while I was composing it). Now that the question has been reopened, the answer may be given here. (This has been edited, but the original letter is revision 1.)


The answer to the question about $S^n$ and their hemispheres $H$ is yes, in a fairly easy way.

The inclusion $i: H \hookrightarrow S^n$ of the northern hemisphere has a retract $r: S^n \to H$ which linearly reflects points on the southern hemisphere across the equatorial hyperplane. Taking $C_1 = S^n$ and $C_0 = H$, and the domain and codomain maps $\partial_0, \partial_1: C_1 \rightrightarrows C_0$ both to be $r$, we may then define the object-to-identity map $u: C_0 \to C_1$ to be $i$, and we get a reflexive graph. Now we define composition to make this to a groupoid. the object $C_2 = C_1 \times_{C_0} C_1$ of composable pairs is just the kernel pair, i.e., the subspace of $C_1 \times C_1$ given by $\{(x, y): r(x) = r(y)\}$. Let $H'$ be the southern hemisphere and let $s: S^n \rightleftarrows H': j$ be the analogous retraction pair for the southern hemisphere, and define $m: C_2 \to C_1$ by $m(x, y) \mapsto i r(x)$ if $x = y$ and $(x, y) \mapsto j s(x)$ if $x \neq y$. It is not hard to check that this defines a continuous map.

The picture is that if you forget the topological structure and look at the underlying discrete groupoid, then there are no morphisms $a \to b$ if $a \neq b$, and the groupoid is a coproduct of groups $\bigsqcup_{a \in H} \hom(a, a)$ where $\hom(a, a)$ is the trivial group if $a$ lies on the equator, and $\hom(a, a)$ is the group $\mathbb{Z}/(2)$ if $a$ is off the equator. As a groupoid this is not connected (since there are no morphisms $a \to b$ between distinct $a, b$), although the total space $C_1$ is a connected space.

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  • $\begingroup$ +1 for your interesting answer. The retracting map is not important in our question. So we do not require that the source and range maps relate to the initial retracting map. We assume that A is retract since this is a necessary condition. $\endgroup$ Commented Aug 19, 2016 at 21:59
  • $\begingroup$ I think, even in the law dimensional case Gl(2,R) and O(2), perhaps the problem is interesting. $\endgroup$ Commented Aug 19, 2016 at 22:04
  • $\begingroup$ Ali, I understand. I've removed editorial comments. $\endgroup$ Commented Aug 19, 2016 at 23:27
  • $\begingroup$ Thanks again for your answer.Regarding the homogenous property of Hom (x,x), when we apply your very interesting point to $Gl(n, $\endgroup$ Commented Aug 21, 2016 at 5:13
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    $\begingroup$ @მამუკაჯიბლაძე There is a smooth groupoid structure as it is written in the answer to the following question:mathoverflow.net/questions/247935/… $\endgroup$ Commented Aug 24, 2016 at 10:50

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