Are all (unbounded) chain complexes over a field bifibrant?












1












$begingroup$


In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where




  • weak equivalences are the quasi-isomorphisms,

  • fibrations are degree-wise surjections, and

  • cofibrations are defined by the left lifting property.


Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.




Question: Is it true that in this case every chain complex is also cofibrant?




I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where




    • weak equivalences are the quasi-isomorphisms,

    • fibrations are degree-wise surjections, and

    • cofibrations are defined by the left lifting property.


    Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.




    Question: Is it true that in this case every chain complex is also cofibrant?




    I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where




      • weak equivalences are the quasi-isomorphisms,

      • fibrations are degree-wise surjections, and

      • cofibrations are defined by the left lifting property.


      Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.




      Question: Is it true that in this case every chain complex is also cofibrant?




      I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?










      share|cite|improve this question









      $endgroup$




      In his paper Homological algebra of homotopy algebras, Hinich defines a model structure on unbounded chain complexes where




      • weak equivalences are the quasi-isomorphisms,

      • fibrations are degree-wise surjections, and

      • cofibrations are defined by the left lifting property.


      Obviously, every object is fibrant. Now assume that we are working over a field $k$ instead of a general ring.




      Question: Is it true that in this case every chain complex is also cofibrant?




      I believe the answer to be yes, and I have an ugly argument that should prove it (even though I still have to check the details). Is there a sly/elegant way to show it?







      homological-algebra model-categories






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      asked Oct 4 '17 at 5:34









      Daniel Robert-NicoudDaniel Robert-Nicoud

      20.5k33797




      20.5k33797






















          2 Answers
          2






          active

          oldest

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          1












          $begingroup$

          Yes.



          Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
          $$dotsto0to kto0todots$$
          and
          $$dotsto0to kstackrel{sim}{to}kto0todots$$
          and their shifts.



          It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good point. Thanks.
            $endgroup$
            – Daniel Robert-Nicoud
            Oct 4 '17 at 12:50



















          1












          $begingroup$

          it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.



          The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Merci Damien ;)
            $endgroup$
            – Daniel Robert-Nicoud
            Jan 12 at 17:13











          Your Answer





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          2 Answers
          2






          active

          oldest

          votes








          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Yes.



          Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
          $$dotsto0to kto0todots$$
          and
          $$dotsto0to kstackrel{sim}{to}kto0todots$$
          and their shifts.



          It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good point. Thanks.
            $endgroup$
            – Daniel Robert-Nicoud
            Oct 4 '17 at 12:50
















          1












          $begingroup$

          Yes.



          Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
          $$dotsto0to kto0todots$$
          and
          $$dotsto0to kstackrel{sim}{to}kto0todots$$
          and their shifts.



          It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Good point. Thanks.
            $endgroup$
            – Daniel Robert-Nicoud
            Oct 4 '17 at 12:50














          1












          1








          1





          $begingroup$

          Yes.



          Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
          $$dotsto0to kto0todots$$
          and
          $$dotsto0to kstackrel{sim}{to}kto0todots$$
          and their shifts.



          It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.






          share|cite|improve this answer









          $endgroup$



          Yes.



          Every (possibly unbounded) chain complex of $k$-vector spaces is a (possibly infinite) direct sum of complexes of the form
          $$dotsto0to kto0todots$$
          and
          $$dotsto0to kstackrel{sim}{to}kto0todots$$
          and their shifts.



          It is easy to show that each of these complexes is cofibrant, and so every complex is cofibrant.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 4 '17 at 12:19









          Jeremy RickardJeremy Rickard

          16.7k11745




          16.7k11745












          • $begingroup$
            Good point. Thanks.
            $endgroup$
            – Daniel Robert-Nicoud
            Oct 4 '17 at 12:50


















          • $begingroup$
            Good point. Thanks.
            $endgroup$
            – Daniel Robert-Nicoud
            Oct 4 '17 at 12:50
















          $begingroup$
          Good point. Thanks.
          $endgroup$
          – Daniel Robert-Nicoud
          Oct 4 '17 at 12:50




          $begingroup$
          Good point. Thanks.
          $endgroup$
          – Daniel Robert-Nicoud
          Oct 4 '17 at 12:50











          1












          $begingroup$

          it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.



          The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Merci Damien ;)
            $endgroup$
            – Daniel Robert-Nicoud
            Jan 12 at 17:13
















          1












          $begingroup$

          it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.



          The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Merci Damien ;)
            $endgroup$
            – Daniel Robert-Nicoud
            Jan 12 at 17:13














          1












          1








          1





          $begingroup$

          it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.



          The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.






          share|cite|improve this answer









          $endgroup$



          it is not difficult to see that when $k$ is a field, then a cofibration is exactly a degree-wise injective map.



          The key lemma is the following: a chain complex whose homology is trivial is homotopic to zero.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 11 at 6:21









          Damien LDamien L

          5,047933




          5,047933












          • $begingroup$
            Merci Damien ;)
            $endgroup$
            – Daniel Robert-Nicoud
            Jan 12 at 17:13


















          • $begingroup$
            Merci Damien ;)
            $endgroup$
            – Daniel Robert-Nicoud
            Jan 12 at 17:13
















          $begingroup$
          Merci Damien ;)
          $endgroup$
          – Daniel Robert-Nicoud
          Jan 12 at 17:13




          $begingroup$
          Merci Damien ;)
          $endgroup$
          – Daniel Robert-Nicoud
          Jan 12 at 17:13


















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