Mixing
Inverting quasi-equivalences between DG categories
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
asked 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
asked 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2 days ago
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
2 days ago
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
asked 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
AGall
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
AGall
132
asked 2 days ago
AGall
132
132
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
AGall is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
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
2 days ago
add a comment |
1 Answer
1
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.)
answered 2 days ago
Kevin Carlson
31.9k23270
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
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♦
yesterday
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.)
answered 2 days ago
Kevin Carlson
31.9k23270
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
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♦
yesterday
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.)
answered 2 days ago
Kevin Carlson
31.9k23270
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
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♦
yesterday
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.)
answered 2 days ago
Kevin Carlson
31.9k23270
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 2 days ago
Kevin Carlson
31.9k23270
answered 2 days ago
Kevin Carlson
31.9k23270
answered 2 days ago
Kevin Carlson
31.9k23270
answered 2 days ago
Kevin Carlson
31.9k23270
31.9k23270
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
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♦
yesterday
add a comment |
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
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♦
yesterday
Thanks a lot! This pretty much settle it!
– AGall
yesterday
Thanks a lot! This pretty much settle it!
– AGall
yesterday
@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
yesterday
@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
yesterday
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♦
yesterday
@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♦
yesterday
add a comment |
AGall is a new contributor. Be nice, and check out our Code of Conduct.
AGall is a new contributor. Be nice, and check out our Code of Conduct.
AGall is a new contributor. Be nice, and check out our Code of Conduct.
AGall is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005475%2finverting-quasi-equivalences-between-dg-categories%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 days ago
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
2 days ago