derived functor that preserves weak equivalences












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Suppose we have a functor $F:Arightarrow B$ between model categories.



1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$



It will be great to see the construction of such derived functor in detail. Thank you.










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$endgroup$

















    3












    $begingroup$


    Suppose we have a functor $F:Arightarrow B$ between model categories.



    1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
    2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$



    It will be great to see the construction of such derived functor in detail. Thank you.










    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      Suppose we have a functor $F:Arightarrow B$ between model categories.



      1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
      2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$



      It will be great to see the construction of such derived functor in detail. Thank you.










      share|cite|improve this question









      $endgroup$




      Suppose we have a functor $F:Arightarrow B$ between model categories.



      1- Assume that F takes weak equivalences to weak equivalences and cofibrations to cofibrations, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$
      2- Assume that F takes weak equivalences to weak equivalences, can we define the derived functor: $$Ho(F): Ho(A)rightarrow Ho(B) $$



      It will be great to see the construction of such derived functor in detail. Thank you.







      reference-request model-categories






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      asked Nov 25 '18 at 18:39









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          $begingroup$

          The answer is yes.



          Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.



          Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
          so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.






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            1 Answer
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            $begingroup$

            The answer is yes.



            Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.



            Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
            so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              The answer is yes.



              Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.



              Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
              so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                The answer is yes.



                Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.



                Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
                so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.






                share|cite|improve this answer









                $endgroup$



                The answer is yes.



                Consider the functor $F:A to Ho(B)$ obtained by composing $F:A to B$ with the localization functor $B to Ho(B)$.



                Then this functor takes weak equivalences in $A$ to isomorphisms in $Ho(B)$,
                so by the universal property of localization, there is a functor $F':Ho(A) to Ho(B)$, such that $F cong F'circ Q$, where $Q:A to Ho(A)$ is the localization functor. This functor $F'$ is your derived functor.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 25 '18 at 19:01









                the Lthe L

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