Is every topological groupoid equivalent to a disjoint union of topological groups?

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It's a fact that any groupoid is equivalent to a disjoint union of (deloopings of) groups. See, e.g. Proposition 4.3 of https://ncatlab.org/nlab/show/groupoid#PropertiesEquivalencesOfGroupoids.



Does this situation carry over to topological groupoids? I.e., is every topological groupoid equivalent to a disjoint union of topological groups?






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    It's a fact that any groupoid is equivalent to a disjoint union of (deloopings of) groups. See, e.g. Proposition 4.3 of https://ncatlab.org/nlab/show/groupoid#PropertiesEquivalencesOfGroupoids.



    Does this situation carry over to topological groupoids? I.e., is every topological groupoid equivalent to a disjoint union of topological groups?






    category-theory topological-groups groupoids







    share|cite|improve this question























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      It's a fact that any groupoid is equivalent to a disjoint union of (deloopings of) groups. See, e.g. Proposition 4.3 of https://ncatlab.org/nlab/show/groupoid#PropertiesEquivalencesOfGroupoids.



      Does this situation carry over to topological groupoids? I.e., is every topological groupoid equivalent to a disjoint union of topological groups?






      category-theory topological-groups groupoids







      share|cite|improve this question













      It's a fact that any groupoid is equivalent to a disjoint union of (deloopings of) groups. See, e.g. Proposition 4.3 of https://ncatlab.org/nlab/show/groupoid#PropertiesEquivalencesOfGroupoids.



      Does this situation carry over to topological groupoids? I.e., is every topological groupoid equivalent to a disjoint union of topological groups?






      category-theory topological-groups groupoids




      category-theory topological-groups groupoids






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      asked 4 hours ago









      Colin

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          No. For instance, any groupoid can be made into a topological groupoid $G$ by giving the spaces of objects and morphisms the indiscrete topology. Any map from $G$ to a disjoint union of topological groups must be constant on objects, since the space of objects of $G$ is connected. It follows that if $G$ has more than one different isomorphism class of objects, then $G$ cannot be equivalent to a disjoint union of topological groups.



          If you want a nice (say, Hausdorff) counterexample, you can take your favorite space $X$ and make a topological groupoid where the space of objects is $X$ and there are no non-identity morphisms (so the space of morphisms is also $X$). By a similar argument to the previous paragraph, this topological groupoid is not equivalent to a disjoint union of topological groups unless $X$ is discrete.






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            No. For instance, any groupoid can be made into a topological groupoid $G$ by giving the spaces of objects and morphisms the indiscrete topology. Any map from $G$ to a disjoint union of topological groups must be constant on objects, since the space of objects of $G$ is connected. It follows that if $G$ has more than one different isomorphism class of objects, then $G$ cannot be equivalent to a disjoint union of topological groups.



            If you want a nice (say, Hausdorff) counterexample, you can take your favorite space $X$ and make a topological groupoid where the space of objects is $X$ and there are no non-identity morphisms (so the space of morphisms is also $X$). By a similar argument to the previous paragraph, this topological groupoid is not equivalent to a disjoint union of topological groups unless $X$ is discrete.






            share|cite|improve this answer
























              up vote
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              No. For instance, any groupoid can be made into a topological groupoid $G$ by giving the spaces of objects and morphisms the indiscrete topology. Any map from $G$ to a disjoint union of topological groups must be constant on objects, since the space of objects of $G$ is connected. It follows that if $G$ has more than one different isomorphism class of objects, then $G$ cannot be equivalent to a disjoint union of topological groups.



              If you want a nice (say, Hausdorff) counterexample, you can take your favorite space $X$ and make a topological groupoid where the space of objects is $X$ and there are no non-identity morphisms (so the space of morphisms is also $X$). By a similar argument to the previous paragraph, this topological groupoid is not equivalent to a disjoint union of topological groups unless $X$ is discrete.






              share|cite|improve this answer






















                up vote
                3
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                up vote
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                down vote



                accepted






                No. For instance, any groupoid can be made into a topological groupoid $G$ by giving the spaces of objects and morphisms the indiscrete topology. Any map from $G$ to a disjoint union of topological groups must be constant on objects, since the space of objects of $G$ is connected. It follows that if $G$ has more than one different isomorphism class of objects, then $G$ cannot be equivalent to a disjoint union of topological groups.



                If you want a nice (say, Hausdorff) counterexample, you can take your favorite space $X$ and make a topological groupoid where the space of objects is $X$ and there are no non-identity morphisms (so the space of morphisms is also $X$). By a similar argument to the previous paragraph, this topological groupoid is not equivalent to a disjoint union of topological groups unless $X$ is discrete.






                share|cite|improve this answer












                No. For instance, any groupoid can be made into a topological groupoid $G$ by giving the spaces of objects and morphisms the indiscrete topology. Any map from $G$ to a disjoint union of topological groups must be constant on objects, since the space of objects of $G$ is connected. It follows that if $G$ has more than one different isomorphism class of objects, then $G$ cannot be equivalent to a disjoint union of topological groups.



                If you want a nice (say, Hausdorff) counterexample, you can take your favorite space $X$ and make a topological groupoid where the space of objects is $X$ and there are no non-identity morphisms (so the space of morphisms is also $X$). By a similar argument to the previous paragraph, this topological groupoid is not equivalent to a disjoint union of topological groups unless $X$ is discrete.







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                share|cite|improve this answer










                answered 4 hours ago









                Eric Wofsey

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