Dual of quasi-isomorphism of chain complexes.











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Let $C_*$, $D_*$ be chain complexes of modules over a ring $R$. Suppose that $fcolon C_* rightarrow D_*$ is a quasi-isomorphism (i.e. an isomorphism in Homology).



I am wondering what conditions are needed (over either $R, C_*$ or $D_*$) so that the dual chain complexes $hom_R(C_*,R)$ and $hom_R(D_*, R)$ are quasi-isomorphic.










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    up vote
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    favorite
    1












    Let $C_*$, $D_*$ be chain complexes of modules over a ring $R$. Suppose that $fcolon C_* rightarrow D_*$ is a quasi-isomorphism (i.e. an isomorphism in Homology).



    I am wondering what conditions are needed (over either $R, C_*$ or $D_*$) so that the dual chain complexes $hom_R(C_*,R)$ and $hom_R(D_*, R)$ are quasi-isomorphic.










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      1





      Let $C_*$, $D_*$ be chain complexes of modules over a ring $R$. Suppose that $fcolon C_* rightarrow D_*$ is a quasi-isomorphism (i.e. an isomorphism in Homology).



      I am wondering what conditions are needed (over either $R, C_*$ or $D_*$) so that the dual chain complexes $hom_R(C_*,R)$ and $hom_R(D_*, R)$ are quasi-isomorphic.










      share|cite|improve this question















      Let $C_*$, $D_*$ be chain complexes of modules over a ring $R$. Suppose that $fcolon C_* rightarrow D_*$ is a quasi-isomorphism (i.e. an isomorphism in Homology).



      I am wondering what conditions are needed (over either $R, C_*$ or $D_*$) so that the dual chain complexes $hom_R(C_*,R)$ and $hom_R(D_*, R)$ are quasi-isomorphic.







      modules homological-algebra abelian-categories






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      edited Nov 17 at 15:20

























      asked Nov 16 at 18:10









      C. Zhihao

      467113




      467113






















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          $newcommand{Hom}{mathrm{Hom}} require{AMScd}$
          Suppose $C$ and $D$ are chain complexes of free $R$-modules, where $R$ is a PID. If $f: C to D$ is a homology isomorphism, then $f^*: Hom(D,R) to Hom(C,R)$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences
          $$ begin{CD} 0 @>>> Ext(H_{ast-1}(D),R) @>>> H^ast(Hom(D,R)) @>>> Hom(H_ast(D),R) @>>>0 \
          @VVV @VVV @VVV @VVV @VVV\
          0 @> >>Ext(H_{ast-1}(C),R)@>>> H^ast(Hom(C,R)) @>>>Hom(H_ast(C),R) @>>>0end{CD}$$

          where the vertical maps are all induced by $f: C to D$. As $f$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.



          For more general $R$, there should be a spectral sequence instead of this short exact sequence giving the same result.






          share|cite|improve this answer





















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            $newcommand{Hom}{mathrm{Hom}} require{AMScd}$
            Suppose $C$ and $D$ are chain complexes of free $R$-modules, where $R$ is a PID. If $f: C to D$ is a homology isomorphism, then $f^*: Hom(D,R) to Hom(C,R)$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences
            $$ begin{CD} 0 @>>> Ext(H_{ast-1}(D),R) @>>> H^ast(Hom(D,R)) @>>> Hom(H_ast(D),R) @>>>0 \
            @VVV @VVV @VVV @VVV @VVV\
            0 @> >>Ext(H_{ast-1}(C),R)@>>> H^ast(Hom(C,R)) @>>>Hom(H_ast(C),R) @>>>0end{CD}$$

            where the vertical maps are all induced by $f: C to D$. As $f$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.



            For more general $R$, there should be a spectral sequence instead of this short exact sequence giving the same result.






            share|cite|improve this answer

























              up vote
              0
              down vote













              $newcommand{Hom}{mathrm{Hom}} require{AMScd}$
              Suppose $C$ and $D$ are chain complexes of free $R$-modules, where $R$ is a PID. If $f: C to D$ is a homology isomorphism, then $f^*: Hom(D,R) to Hom(C,R)$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences
              $$ begin{CD} 0 @>>> Ext(H_{ast-1}(D),R) @>>> H^ast(Hom(D,R)) @>>> Hom(H_ast(D),R) @>>>0 \
              @VVV @VVV @VVV @VVV @VVV\
              0 @> >>Ext(H_{ast-1}(C),R)@>>> H^ast(Hom(C,R)) @>>>Hom(H_ast(C),R) @>>>0end{CD}$$

              where the vertical maps are all induced by $f: C to D$. As $f$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.



              For more general $R$, there should be a spectral sequence instead of this short exact sequence giving the same result.






              share|cite|improve this answer























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                up vote
                0
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                $newcommand{Hom}{mathrm{Hom}} require{AMScd}$
                Suppose $C$ and $D$ are chain complexes of free $R$-modules, where $R$ is a PID. If $f: C to D$ is a homology isomorphism, then $f^*: Hom(D,R) to Hom(C,R)$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences
                $$ begin{CD} 0 @>>> Ext(H_{ast-1}(D),R) @>>> H^ast(Hom(D,R)) @>>> Hom(H_ast(D),R) @>>>0 \
                @VVV @VVV @VVV @VVV @VVV\
                0 @> >>Ext(H_{ast-1}(C),R)@>>> H^ast(Hom(C,R)) @>>>Hom(H_ast(C),R) @>>>0end{CD}$$

                where the vertical maps are all induced by $f: C to D$. As $f$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.



                For more general $R$, there should be a spectral sequence instead of this short exact sequence giving the same result.






                share|cite|improve this answer












                $newcommand{Hom}{mathrm{Hom}} require{AMScd}$
                Suppose $C$ and $D$ are chain complexes of free $R$-modules, where $R$ is a PID. If $f: C to D$ is a homology isomorphism, then $f^*: Hom(D,R) to Hom(C,R)$ is as well. This follows from the naturality of the Universal Coefficient Theorem, as there is a commuting diagram of exact sequences
                $$ begin{CD} 0 @>>> Ext(H_{ast-1}(D),R) @>>> H^ast(Hom(D,R)) @>>> Hom(H_ast(D),R) @>>>0 \
                @VVV @VVV @VVV @VVV @VVV\
                0 @> >>Ext(H_{ast-1}(C),R)@>>> H^ast(Hom(C,R)) @>>>Hom(H_ast(C),R) @>>>0end{CD}$$

                where the vertical maps are all induced by $f: C to D$. As $f$ is a homology isomorphism, all maps except the middle are isomorphisms. Hence, the Five Lemma gives that the middle map is an isomorphism as well.



                For more general $R$, there should be a spectral sequence instead of this short exact sequence giving the same result.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 17 at 20:38









                Joshua Mundinger

                2,361926




                2,361926






























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