Quotient map between homotopy equivalent spaces












1












$begingroup$


$require{AMScd}$
Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
begin{CD}
X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
end{CD}
such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
Is it true that $q$ is then a homotopy equivalence?



Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?






general-topology







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    $require{AMScd}$
    Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
    begin{CD}
    X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
    end{CD}
    such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
    Is it true that $q$ is then a homotopy equivalence?



    Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?






    general-topology







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      $require{AMScd}$
      Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
      begin{CD}
      X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
      end{CD}
      such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
      Is it true that $q$ is then a homotopy equivalence?



      Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?






      general-topology







      share|cite|improve this question









      $endgroup$




      $require{AMScd}$
      Let $X$ be some top. space and $X/{raise.17exhbox{$scriptstylesim$}} $ a quotient space with quotient map $q$:
      begin{CD}
      X @>q>> X/{raise.17exhbox{$scriptstylesim$}}
      end{CD}
      such that X is homotopy equivalent (or homeomorphic if that makes a difference) to $X/{raise.17exhbox{$scriptstylesim$}} $.
      Is it true that $q$ is then a homotopy equivalence?



      Thinking about 'nice' examples this seemed reasonable, at least I didn't see an example where it goes wrong. Trying to prove it though, there didn't seem to be any reasoning why $q$ should be homotopic to the given homotopy equivalence. So I assume this doesn't hold in general? Is it perhaps true for 'nice' spaces such as CW-complexes?






      general-topology




      general-topology






      share|cite|improve this question













      share|cite|improve this question











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      asked Dec 17 '18 at 16:47









      BertrandBertrand

      685




      685






















          2 Answers
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          active

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          $begingroup$

          There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).



          Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.



          Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:




            If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.







            share|cite|improve this answer









            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

              oldest

              votes









              3












              $begingroup$

              There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).



              Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.



              Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).



                Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.



                Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).



                  Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.



                  Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.






                  share|cite|improve this answer











                  $endgroup$



                  There should be enough counterexamples to this when your space is "big enough" (but otherwise well-behaved).



                  Consider, for example, the discrete countable space $mathbb{N}$. Given any partition of $mathbb{N}$ into finite subsets $mathbb{N}=coprod_{n in mathbb{N}}F_n$, there is the quotient $X/sim$ where $sim$ is the equivalence such that $F_n$'s are its equivalence classes, i.e. $x sim y$ iff $x, y in F_n$for some $n$. Then $X$ and $X/sim$ are homeomorphic (for example by $n mapsto F_n,$ both are countable discrete spaces). But the quotient map is not a homotopy equivalence, because it does not induce bijection on the set of path-components, which are singletons in both cases, unless all the sets $F_n$ are themselves singletons.



                  Similarly, you can take e.g. $X=mathbb{C}setminus mathbb{Z}$, and the quotient $X/sim$ obtained by shrinking ${z in mathbb{C};|; 0neq |z| leqfrac{1}{2}}$ into a point. Then again, $X/sim$ is homeomorphic to $X$ (both are a plane with countable set of isolated punctures), but the quotient map does not induce isomorphism on fundamental groups, because the non-trivial loop that goes once around the origin in X becomes trivial after applying $q$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Dec 17 '18 at 23:11

























                  answered Dec 17 '18 at 22:21









                  Pavel ČoupekPavel Čoupek

                  4,45611126




                  4,45611126























                      0












                      $begingroup$

                      Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:




                      If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.







                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:




                        If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.







                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:




                          If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.







                          share|cite|improve this answer









                          $endgroup$



                          Here is (only) a partial answer regarding CW complexes. It's proposition 0.17 in Hatcher's book:




                          If $(X,A)$ is a $CW$ pair consisting of a $CW$ complex $X$ and a contratible subcomplex $A,$ then the quotient map $Xto X/A$ is a homotopy equivalence.








                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Dec 17 '18 at 17:00









                          positrón0802positrón0802

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                          4,353520






























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