Is quasi-isomorpism of $A_{infty}$ algebras invertible












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Let $A_1$ and $A_2$ be two $A_{infty}$-algebras over a field $k$. Suppose $f={f_i} : A_1 to A_2$ is $A_{infty}$ quasi-isomorphism of $A_{infty}$-algebras that is a map of $A_{infty}$-algebras such that $f_1$ is quasi-isomorphism. Under such assumptions is $f$ invertible? More precisely, is there an $A_{infty}$-map $g: A_2 to A_1$ such that compositions $fcirc g$ and $g circ f$ induce identity map on cohomology?






reference-request homological-algebra







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$endgroup$












  • $begingroup$
    Yes. The proof is similar to the case of formal power series.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:15












  • $begingroup$
    Thanks! Do you know any references for this fact?
    $endgroup$
    – Alex
    Jan 25 at 9:16






  • 1




    $begingroup$
    This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:26


















0












$begingroup$


Let $A_1$ and $A_2$ be two $A_{infty}$-algebras over a field $k$. Suppose $f={f_i} : A_1 to A_2$ is $A_{infty}$ quasi-isomorphism of $A_{infty}$-algebras that is a map of $A_{infty}$-algebras such that $f_1$ is quasi-isomorphism. Under such assumptions is $f$ invertible? More precisely, is there an $A_{infty}$-map $g: A_2 to A_1$ such that compositions $fcirc g$ and $g circ f$ induce identity map on cohomology?






reference-request homological-algebra







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes. The proof is similar to the case of formal power series.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:15












  • $begingroup$
    Thanks! Do you know any references for this fact?
    $endgroup$
    – Alex
    Jan 25 at 9:16






  • 1




    $begingroup$
    This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:26
















0












0








0





$begingroup$


Let $A_1$ and $A_2$ be two $A_{infty}$-algebras over a field $k$. Suppose $f={f_i} : A_1 to A_2$ is $A_{infty}$ quasi-isomorphism of $A_{infty}$-algebras that is a map of $A_{infty}$-algebras such that $f_1$ is quasi-isomorphism. Under such assumptions is $f$ invertible? More precisely, is there an $A_{infty}$-map $g: A_2 to A_1$ such that compositions $fcirc g$ and $g circ f$ induce identity map on cohomology?






reference-request homological-algebra







share|cite|improve this question









$endgroup$




Let $A_1$ and $A_2$ be two $A_{infty}$-algebras over a field $k$. Suppose $f={f_i} : A_1 to A_2$ is $A_{infty}$ quasi-isomorphism of $A_{infty}$-algebras that is a map of $A_{infty}$-algebras such that $f_1$ is quasi-isomorphism. Under such assumptions is $f$ invertible? More precisely, is there an $A_{infty}$-map $g: A_2 to A_1$ such that compositions $fcirc g$ and $g circ f$ induce identity map on cohomology?






reference-request homological-algebra




reference-request homological-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 25 at 9:12









AlexAlex

2,8941128




2,8941128












  • $begingroup$
    Yes. The proof is similar to the case of formal power series.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:15












  • $begingroup$
    Thanks! Do you know any references for this fact?
    $endgroup$
    – Alex
    Jan 25 at 9:16






  • 1




    $begingroup$
    This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:26




















  • $begingroup$
    Yes. The proof is similar to the case of formal power series.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:15












  • $begingroup$
    Thanks! Do you know any references for this fact?
    $endgroup$
    – Alex
    Jan 25 at 9:16






  • 1




    $begingroup$
    This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
    $endgroup$
    – Pedro Tamaroff♦
    Jan 25 at 9:26


















$begingroup$
Yes. The proof is similar to the case of formal power series.
$endgroup$
– Pedro Tamaroff♦
Jan 25 at 9:15






$begingroup$
Yes. The proof is similar to the case of formal power series.
$endgroup$
– Pedro Tamaroff♦
Jan 25 at 9:15














$begingroup$
Thanks! Do you know any references for this fact?
$endgroup$
– Alex
Jan 25 at 9:16




$begingroup$
Thanks! Do you know any references for this fact?
$endgroup$
– Alex
Jan 25 at 9:16




1




1




$begingroup$
This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
$endgroup$
– Pedro Tamaroff♦
Jan 25 at 9:26






$begingroup$
This is in the thesis of Alain Proute, Theorem 4.27. The computations can get nasty.
$endgroup$
– Pedro Tamaroff♦
Jan 25 at 9:26












1 Answer
1






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oldest

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1












$begingroup$

In Kellers "Introduction to $A_infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ]
In particular you find an $g$ such that $f circ g sim id sim g circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.



The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you arw owrking over a field).



However that might need that $k$ is a field, but I think you should be happy with that.



This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.






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    $begingroup$

    In Kellers "Introduction to $A_infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ]
    In particular you find an $g$ such that $f circ g sim id sim g circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.



    The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you arw owrking over a field).



    However that might need that $k$ is a field, but I think you should be happy with that.



    This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      In Kellers "Introduction to $A_infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ]
      In particular you find an $g$ such that $f circ g sim id sim g circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.



      The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you arw owrking over a field).



      However that might need that $k$ is a field, but I think you should be happy with that.



      This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        In Kellers "Introduction to $A_infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ]
        In particular you find an $g$ such that $f circ g sim id sim g circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.



        The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you arw owrking over a field).



        However that might need that $k$ is a field, but I think you should be happy with that.



        This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.






        share|cite|improve this answer









        $endgroup$



        In Kellers "Introduction to $A_infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ]
        In particular you find an $g$ such that $f circ g sim id sim g circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.



        The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you arw owrking over a field).



        However that might need that $k$ is a field, but I think you should be happy with that.



        This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 25 at 10:36









        EnkiduEnkidu

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