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Endomorphisms in the derived category
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(Apologies if this question is trivial, but I'm way outside my area here.)
Let $R$ be a commutative ring, $C^bullet(R)$ the category of complexes of $R$-modules, and $D^bullet(R)$ its derived category. Suppose we have an object $X in operatornameObj D^bullet(R)$, and an endomorphism $t in operatornameEnd_D^bullet(R) X$. By definition, this just means that we can write $t$ as a formal fraction $s^-1 f$, where $f: X^bulletto Y^bullet$ is a map of complexes and $s: X^bullet to Y^bullet$ is a quasi-isomorphism.
Can we always arrange that $X^bullet = Y^bullet$? That is, do endomorphisms in the derived category always lift to endomorphisms of some (or even any) representing complex?
Something of this kind seems to be used at the bottom of page 10 of this paper by Khare and Thorne, and I'm trying to work out why this is true.
ct.category-theory derived-categories
asked 2 hours ago
David Loeffler
19.1k146113
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up vote
3
down vote
favorite
(Apologies if this question is trivial, but I'm way outside my area here.)
Let $R$ be a commutative ring, $C^bullet(R)$ the category of complexes of $R$-modules, and $D^bullet(R)$ its derived category. Suppose we have an object $X in operatornameObj D^bullet(R)$, and an endomorphism $t in operatornameEnd_D^bullet(R) X$. By definition, this just means that we can write $t$ as a formal fraction $s^-1 f$, where $f: X^bulletto Y^bullet$ is a map of complexes and $s: X^bullet to Y^bullet$ is a quasi-isomorphism.
Can we always arrange that $X^bullet = Y^bullet$? That is, do endomorphisms in the derived category always lift to endomorphisms of some (or even any) representing complex?
Something of this kind seems to be used at the bottom of page 10 of this paper by Khare and Thorne, and I'm trying to work out why this is true.
ct.category-theory derived-categories
asked 2 hours ago
David Loeffler
19.1k146113
1
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
(Apologies if this question is trivial, but I'm way outside my area here.)
Let $R$ be a commutative ring, $C^bullet(R)$ the category of complexes of $R$-modules, and $D^bullet(R)$ its derived category. Suppose we have an object $X in operatornameObj D^bullet(R)$, and an endomorphism $t in operatornameEnd_D^bullet(R) X$. By definition, this just means that we can write $t$ as a formal fraction $s^-1 f$, where $f: X^bulletto Y^bullet$ is a map of complexes and $s: X^bullet to Y^bullet$ is a quasi-isomorphism.
Can we always arrange that $X^bullet = Y^bullet$? That is, do endomorphisms in the derived category always lift to endomorphisms of some (or even any) representing complex?
Something of this kind seems to be used at the bottom of page 10 of this paper by Khare and Thorne, and I'm trying to work out why this is true.
ct.category-theory derived-categories
asked 2 hours ago
David Loeffler
19.1k146113
(Apologies if this question is trivial, but I'm way outside my area here.)
Let $R$ be a commutative ring, $C^bullet(R)$ the category of complexes of $R$-modules, and $D^bullet(R)$ its derived category. Suppose we have an object $X in operatornameObj D^bullet(R)$, and an endomorphism $t in operatornameEnd_D^bullet(R) X$. By definition, this just means that we can write $t$ as a formal fraction $s^-1 f$, where $f: X^bulletto Y^bullet$ is a map of complexes and $s: X^bullet to Y^bullet$ is a quasi-isomorphism.
Can we always arrange that $X^bullet = Y^bullet$? That is, do endomorphisms in the derived category always lift to endomorphisms of some (or even any) representing complex?
Something of this kind seems to be used at the bottom of page 10 of this paper by Khare and Thorne, and I'm trying to work out why this is true.
ct.category-theory derived-categories
ct.category-theory derived-categories
asked 2 hours ago
David Loeffler
19.1k146113
asked 2 hours ago
David Loeffler
19.1k146113
asked 2 hours ago
David Loeffler
19.1k146113
asked 2 hours ago
David Loeffler
19.1k146113
asked 2 hours ago
David Loeffler
19.1k146113
19.1k146113
1
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago
add a comment |Â
1
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago
1
1
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago
add a comment |Â
1 Answer
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It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:
Let $X_bullet$ be a complex. There exist a complex $X'_bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'bulletto X_bullet$, such that every endo-morphism of $X_bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_bullet to X_bullet$, and it has this property.
answered 1 hour ago
S. carmeli
1,990318
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:
Let $X_bullet$ be a complex. There exist a complex $X'_bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'bulletto X_bullet$, such that every endo-morphism of $X_bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_bullet to X_bullet$, and it has this property.
answered 1 hour ago
S. carmeli
1,990318
add a comment |Â
up vote
4
down vote
It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:
Let $X_bullet$ be a complex. There exist a complex $X'_bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'bulletto X_bullet$, such that every endo-morphism of $X_bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_bullet to X_bullet$, and it has this property.
answered 1 hour ago
S. carmeli
1,990318
add a comment |Â
up vote
4
down vote
up vote
4
down vote
It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:
Let $X_bullet$ be a complex. There exist a complex $X'_bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'bulletto X_bullet$, such that every endo-morphism of $X_bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_bullet to X_bullet$, and it has this property.
answered 1 hour ago
S. carmeli
1,990318
It slightly depends what you mean by "arrange that $X=Y$". As is the statement is not true. There are complexes with endomorphisms which are not realizable. However, what is true is the following:
Let $X_bullet$ be a complex. There exist a complex $X'_bullet$ and a morphism of complexes which is a quasi-isomorphism $ X'bulletto X_bullet$, such that every endo-morphism of $X_bullet'$ can be realized as an endo-morphism in the complex category. To see this, choose a projective resolution $X'_bullet to X_bullet$, and it has this property.
answered 1 hour ago
S. carmeli
1,990318
answered 1 hour ago
S. carmeli
1,990318
answered 1 hour ago
S. carmeli
1,990318
answered 1 hour ago
S. carmeli
1,990318
1,990318
add a comment |Â
add a comment |Â
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1
If I read it correctly, this follows by $C$ being a bounded perfect complex (i.e. levelwise projective). So any quasiiso between them is already a homotopy equivalence, and so it has a representative. There are quite some reasons you can argue that, either by homological algebra, model categories...
– Enkidu
2 hours ago