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.
modules homological-algebra abelian-categories
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up vote
2
down vote
favorite
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
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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
modules homological-algebra abelian-categories
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.
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
add a comment |
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.
add a comment |
up vote
0
down vote
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.
$newcommand{Hom}{mathrm{Hom}} require{AMScd}$
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.
answered Nov 17 at 20:38
Joshua Mundinger
2,361926
2,361926
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