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 $X in mathrm{Ob}(X)$.










share|cite|improve this question






















  • 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















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 $X in mathrm{Ob}(X)$.










share|cite|improve this question






















  • 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













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 $X in mathrm{Ob}(X)$.










share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • 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










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.)






share|cite|improve this answer





















  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















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

























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.)






share|cite|improve this answer





















  • 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















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.)






share|cite|improve this answer





















  • 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













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.)






share|cite|improve this answer












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.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 20 at 5:43









Kevin Carlson

32.3k23270




32.3k23270












  • 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










  • @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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














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





















































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







Popular posts from this blog

Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

ComboBox Display Member on multiple fields

Is it possible to collect Nectar points via Trainline?