$K_star$ and $L_star$ are homotopic as chain complex. Is it true that...












1












$begingroup$


Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.



$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.






abstract-algebra homological-algebra







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$endgroup$












  • $begingroup$
    What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
    $endgroup$
    – Eric Wofsey
    Dec 19 '18 at 19:33






  • 3




    $begingroup$
    Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:33








  • 1




    $begingroup$
    @EricWofsey I think OP means the two complexes are chain homotopy equivalent
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:35










  • $begingroup$
    @EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
    $endgroup$
    – user45765
    Dec 19 '18 at 19:35
















1












$begingroup$


Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.



$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.






abstract-algebra homological-algebra







share|cite|improve this question











$endgroup$












  • $begingroup$
    What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
    $endgroup$
    – Eric Wofsey
    Dec 19 '18 at 19:33






  • 3




    $begingroup$
    Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:33








  • 1




    $begingroup$
    @EricWofsey I think OP means the two complexes are chain homotopy equivalent
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:35










  • $begingroup$
    @EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
    $endgroup$
    – user45765
    Dec 19 '18 at 19:35














1












1








1


1



$begingroup$


Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.



$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.






abstract-algebra homological-algebra







share|cite|improve this question











$endgroup$




Consider $K_star,L_star$ 2 chain complexes and suppose they are homotopic say $f:Kto L$.



$textbf{Q:}$ Consider $operatorname{Hom}(K_star, L_star)$ complex and take its homology. It is clear that $[f]=0in H_0(operatorname{Hom}_star(K,L))$ as $f$ is a homotopy(i.e. It is boundary of $operatorname{Hom}_1(K_star,L_star)$). However, do I even know or can say $H_star(operatorname{Hom}(K,L))=0$? Furthermore, when do I know $H_star(operatorname{Hom}(K,L))=0$. I doubt I can even say $H_0(operatorname{Hom}(K,L))=0$ in general as it might be very large.






abstract-algebra homological-algebra




abstract-algebra homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 19:33









Eric Wofsey

192k14217351




192k14217351










asked Dec 19 '18 at 19:17









user45765user45765

2,6882724




2,6882724












  • $begingroup$
    What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
    $endgroup$
    – Eric Wofsey
    Dec 19 '18 at 19:33






  • 3




    $begingroup$
    Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:33








  • 1




    $begingroup$
    @EricWofsey I think OP means the two complexes are chain homotopy equivalent
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:35










  • $begingroup$
    @EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
    $endgroup$
    – user45765
    Dec 19 '18 at 19:35


















  • $begingroup$
    What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
    $endgroup$
    – Eric Wofsey
    Dec 19 '18 at 19:33






  • 3




    $begingroup$
    Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:33








  • 1




    $begingroup$
    @EricWofsey I think OP means the two complexes are chain homotopy equivalent
    $endgroup$
    – leibnewtz
    Dec 19 '18 at 19:35










  • $begingroup$
    @EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
    $endgroup$
    – user45765
    Dec 19 '18 at 19:35
















$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33




$begingroup$
What do you mean when you say two chain complexes are "homotopic"? That is not standard terminology.
$endgroup$
– Eric Wofsey
Dec 19 '18 at 19:33




3




3




$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33






$begingroup$
Let $K=L=mathbb{Z}$ concentrated in degree zero. By freeness, the hom complex will be the ordinary hom module, and its homology will be itself
$endgroup$
– leibnewtz
Dec 19 '18 at 19:33






1




1




$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35




$begingroup$
@EricWofsey I think OP means the two complexes are chain homotopy equivalent
$endgroup$
– leibnewtz
Dec 19 '18 at 19:35












$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35




$begingroup$
@EricWofsey Sorry for confusion. I really should say they are equivalent by a map. Pick a map $f:Kto L$ and suppose there is another map $g:Lto K$ s.t. $f,g$ induce homotopy equivalence of chain complexes.
$endgroup$
– user45765
Dec 19 '18 at 19:35










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$begingroup$

It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
$$H_0(underline{Hom}(L,K)) = [L,K]$$
So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:





  1. $H_*(underline{Hom}(L,K)) = 0$,


  2. $H_0(underline{Hom}(L,K)) = 0$,


  3. $L$ is contractible, i.e. $text{id}_L simeq 0$.



In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.






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    $begingroup$

    It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
    $$H_0(underline{Hom}(L,K)) = [L,K]$$
    So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:





    1. $H_*(underline{Hom}(L,K)) = 0$,


    2. $H_0(underline{Hom}(L,K)) = 0$,


    3. $L$ is contractible, i.e. $text{id}_L simeq 0$.



    In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
      $$H_0(underline{Hom}(L,K)) = [L,K]$$
      So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:





      1. $H_*(underline{Hom}(L,K)) = 0$,


      2. $H_0(underline{Hom}(L,K)) = 0$,


      3. $L$ is contractible, i.e. $text{id}_L simeq 0$.



      In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
        $$H_0(underline{Hom}(L,K)) = [L,K]$$
        So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:





        1. $H_*(underline{Hom}(L,K)) = 0$,


        2. $H_0(underline{Hom}(L,K)) = 0$,


        3. $L$ is contractible, i.e. $text{id}_L simeq 0$.



        In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.






        share|cite|improve this answer









        $endgroup$



        It's a general fact that the 0th homology of the internal hom of chain complexes equals the homotopy classes of chain maps:
        $$H_0(underline{Hom}(L,K)) = [L,K]$$
        So, $[f] = 0$ iff $f$ is nulhomotopic. If you are assuming $f$ is an equivalence then $[f] = 0 $ implies $L,K$ are contractible. Similarly, if $Lsimeq K$, then the following are equivalent:





        1. $H_*(underline{Hom}(L,K)) = 0$,


        2. $H_0(underline{Hom}(L,K)) = 0$,


        3. $L$ is contractible, i.e. $text{id}_L simeq 0$.



        In particular, if $f$ is a homotopy equivalence it does not follow that $[f] = 0$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 21 '18 at 11:44









        BenBen

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