Inverting quasi-equivalences between DG categories
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I am recently trying to learn the language of DG categories and I have a question concerning the notion of quasi-equivalence.
According to the definition, which you can find for instance on Keller's paper "On differential graded categories", for a given DG functor $F colon mathcal{A} rightarrow mathcal{B}$ to be a quasi-equivalence means that for all $X,Y in mathrm{Ob}(mathcal{A})$ the induced map
$$
F_{X,Y} colon mathrm{Hom}_{mathcal{A}}^{bullet}(X,Y) rightarrow mathrm{Hom}_{mathcal{B}}^{bullet}(F(X),F(Y))
$$
of chain complexes is a quasi-isomorphism, and moreover that the induced functor
$$
H^0(F) colon H^0(mathcal{A}) rightarrow H^0(mathcal{B})
$$
on the level of categories is essentially surjective.
Now, in ordinary category theory, a given functor $F colon mathcal{A} rightarrow mathcal{B}$ is an equivalence of categories if and only if one can find another functor $G colon mathcal{B} rightarrow mathcal{A}$ such that $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$. This so-called ``quasi-inverse'' turns out to be unique up to natural equivalence.
My question is the following: Given a quasi-equivalence $F colon mathcal{A} rightarrow mathcal{B}$ is it possible to find a quasi-equivalence $G colon mathcal{B} rightarrow mathcal{A}$ together with DG natural isomorphisms $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$?
Here, by a DG natural isomorphism between two DG functors $F,G colon mathcal{A} rightarrow mathcal{B}$, I mean a DG natural transformation $varphi colon F Rightarrow G$ of degree $0$, as defined for instance in Genovese's paper "The uniqueness problem of dg-lifts and Fourier-Mukai kernels", such that $varphi(X)$ is an isomorphism for all $X in mathrm{Ob}(X)$.
category-theory homological-algebra
add a comment |
up vote
2
down vote
favorite
I am recently trying to learn the language of DG categories and I have a question concerning the notion of quasi-equivalence.
According to the definition, which you can find for instance on Keller's paper "On differential graded categories", for a given DG functor $F colon mathcal{A} rightarrow mathcal{B}$ to be a quasi-equivalence means that for all $X,Y in mathrm{Ob}(mathcal{A})$ the induced map
$$
F_{X,Y} colon mathrm{Hom}_{mathcal{A}}^{bullet}(X,Y) rightarrow mathrm{Hom}_{mathcal{B}}^{bullet}(F(X),F(Y))
$$
of chain complexes is a quasi-isomorphism, and moreover that the induced functor
$$
H^0(F) colon H^0(mathcal{A}) rightarrow H^0(mathcal{B})
$$
on the level of categories is essentially surjective.
Now, in ordinary category theory, a given functor $F colon mathcal{A} rightarrow mathcal{B}$ is an equivalence of categories if and only if one can find another functor $G colon mathcal{B} rightarrow mathcal{A}$ such that $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$. This so-called ``quasi-inverse'' turns out to be unique up to natural equivalence.
My question is the following: Given a quasi-equivalence $F colon mathcal{A} rightarrow mathcal{B}$ is it possible to find a quasi-equivalence $G colon mathcal{B} rightarrow mathcal{A}$ together with DG natural isomorphisms $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$?
Here, by a DG natural isomorphism between two DG functors $F,G colon mathcal{A} rightarrow mathcal{B}$, I mean a DG natural transformation $varphi colon F Rightarrow G$ of degree $0$, as defined for instance in Genovese's paper "The uniqueness problem of dg-lifts and Fourier-Mukai kernels", such that $varphi(X)$ is an isomorphism for all $X in mathrm{Ob}(X)$.
category-theory homological-algebra
In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am recently trying to learn the language of DG categories and I have a question concerning the notion of quasi-equivalence.
According to the definition, which you can find for instance on Keller's paper "On differential graded categories", for a given DG functor $F colon mathcal{A} rightarrow mathcal{B}$ to be a quasi-equivalence means that for all $X,Y in mathrm{Ob}(mathcal{A})$ the induced map
$$
F_{X,Y} colon mathrm{Hom}_{mathcal{A}}^{bullet}(X,Y) rightarrow mathrm{Hom}_{mathcal{B}}^{bullet}(F(X),F(Y))
$$
of chain complexes is a quasi-isomorphism, and moreover that the induced functor
$$
H^0(F) colon H^0(mathcal{A}) rightarrow H^0(mathcal{B})
$$
on the level of categories is essentially surjective.
Now, in ordinary category theory, a given functor $F colon mathcal{A} rightarrow mathcal{B}$ is an equivalence of categories if and only if one can find another functor $G colon mathcal{B} rightarrow mathcal{A}$ such that $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$. This so-called ``quasi-inverse'' turns out to be unique up to natural equivalence.
My question is the following: Given a quasi-equivalence $F colon mathcal{A} rightarrow mathcal{B}$ is it possible to find a quasi-equivalence $G colon mathcal{B} rightarrow mathcal{A}$ together with DG natural isomorphisms $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$?
Here, by a DG natural isomorphism between two DG functors $F,G colon mathcal{A} rightarrow mathcal{B}$, I mean a DG natural transformation $varphi colon F Rightarrow G$ of degree $0$, as defined for instance in Genovese's paper "The uniqueness problem of dg-lifts and Fourier-Mukai kernels", such that $varphi(X)$ is an isomorphism for all $X in mathrm{Ob}(X)$.
category-theory homological-algebra
I am recently trying to learn the language of DG categories and I have a question concerning the notion of quasi-equivalence.
According to the definition, which you can find for instance on Keller's paper "On differential graded categories", for a given DG functor $F colon mathcal{A} rightarrow mathcal{B}$ to be a quasi-equivalence means that for all $X,Y in mathrm{Ob}(mathcal{A})$ the induced map
$$
F_{X,Y} colon mathrm{Hom}_{mathcal{A}}^{bullet}(X,Y) rightarrow mathrm{Hom}_{mathcal{B}}^{bullet}(F(X),F(Y))
$$
of chain complexes is a quasi-isomorphism, and moreover that the induced functor
$$
H^0(F) colon H^0(mathcal{A}) rightarrow H^0(mathcal{B})
$$
on the level of categories is essentially surjective.
Now, in ordinary category theory, a given functor $F colon mathcal{A} rightarrow mathcal{B}$ is an equivalence of categories if and only if one can find another functor $G colon mathcal{B} rightarrow mathcal{A}$ such that $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$. This so-called ``quasi-inverse'' turns out to be unique up to natural equivalence.
My question is the following: Given a quasi-equivalence $F colon mathcal{A} rightarrow mathcal{B}$ is it possible to find a quasi-equivalence $G colon mathcal{B} rightarrow mathcal{A}$ together with DG natural isomorphisms $F circ G cong mathrm{id}_{mathcal{B}}$ and $G circ F cong mathrm{id}_{mathcal{A}}$?
Here, by a DG natural isomorphism between two DG functors $F,G colon mathcal{A} rightarrow mathcal{B}$, I mean a DG natural transformation $varphi colon F Rightarrow G$ of degree $0$, as defined for instance in Genovese's paper "The uniqueness problem of dg-lifts and Fourier-Mukai kernels", such that $varphi(X)$ is an isomorphism for all $X in mathrm{Ob}(X)$.
category-theory homological-algebra
category-theory homological-algebra
asked Nov 19 at 20:23
AGall
132
132
In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24
add a comment |
In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24
In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24
add a comment |
1 Answer
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LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=mathbb{Z}to mathbb{Z}$ while $B(0,1)=mathbb{Z}/2$. Then there is a quasi-equivalence $Ato B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=mathbb{Z}to mathbb{Z}$ while $B(0,1)=mathbb{Z}/2$. Then there is a quasi-equivalence $Ato B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
add a comment |
up vote
1
down vote
accepted
LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=mathbb{Z}to mathbb{Z}$ while $B(0,1)=mathbb{Z}/2$. Then there is a quasi-equivalence $Ato B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=mathbb{Z}to mathbb{Z}$ while $B(0,1)=mathbb{Z}/2$. Then there is a quasi-equivalence $Ato B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)
LPK's example in the comments can be upgraded to a counterexample to exactly this situation. Consider two DG-categories $A,B$, both with two objects $0,1$ and morphisms $A(0,0)=A(1,1)=B(0,0)=B(1,1)=mathbb{Z}$, $A(1,0)=B(1,0)=0$, and finally $A(0,1)=mathbb{Z}to mathbb{Z}$ while $B(0,1)=mathbb{Z}/2$. Then there is a quasi-equivalence $Ato B$ which is the identity on objects and has the usual quasi-isomorphism $A(0,1)to B(0,1)$ as its only nontrivial morphism action. This is not invertible since $A(0,1)to B(0,1)$ is not. The problem is that $A$ is not bifibrant as a DG category. This is the model category theoretic condition that, generally, realizes weak equivalences (such quasi-equivalences) as homotopy equivalences-legitimately weakly invertible morphisms. The bifibrant DG-categories are difficult to describe fully explicitly, but at the very least the hom-complexes must be levelwise projective (or perhaps injective, in a different model structure.)
answered Nov 20 at 5:43
Kevin Carlson
32.2k23270
32.2k23270
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
add a comment |
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
Thanks a lot! This pretty much settle it!
– AGall
Nov 20 at 14:30
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
@AGall Thanks! If you're satisfied with an answer, you can click the check mark in the upper left to accept it-the question then gets removed from the list of unsettled questions.
– Kevin Carlson
Nov 20 at 17:12
1
1
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
@AGall You can find other examples in Töen's Lectures on dg categories. One can usually upgrade, like Kevin did, a counterexample on the dg algebra level or chain complex level to one of dg categories.
– Pedro Tamaroff♦
Nov 20 at 23:44
add a comment |
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In general, I think the answer should be no. Think about it like this: if you are given a morphism of chain complexes which is a quasi-isomorphism, you have an inverse map on the homologies, but these inverse maps in general do not lift to a map of complexes.
– LPK
Nov 19 at 21:19
An example of this is the following: consider the complex $C$ with $mathbb{Z}$ in degree 0 and 1, with the map "multiply by two" between them. Consider another complex $D$ with $mathbb{Z}/2$ in degree 1 and zeros elsewhere. There is a chain map $f_{ast}: C rightarrow D$ where $f_{1}$ is the quotient $mathbb{Z} rightarrow mathbb{Z}/2$. This induces a quasi-isomorphism, but there is no map $mathbb{Z}/2 rightarrow mathbb{Z}$.
– LPK
Nov 19 at 21:24