simplicial objects in a model category












4












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Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?






ct.category-theory homotopy-theory model-categories







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Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    4












    $begingroup$


    Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
    We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?






    ct.category-theory homotopy-theory model-categories







    share|cite|improve this question







    New contributor




    Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      3



      $begingroup$


      Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
      We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?






      ct.category-theory homotopy-theory model-categories







      share|cite|improve this question







      New contributor




      Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      Suppose that we have a (combinatorial if necessary) model category $M$, and let $F:Delta^{op}rightarrow M$ a simplicial object in $M$, such that for any natural number $n$, $F([n])$ is a fibrant object in $M$.
      We define a new object $X= colim_{n} F([n]) $. Is it true that $X$ is a fibrant object ?






      ct.category-theory homotopy-theory model-categories




      ct.category-theory homotopy-theory model-categories






      share|cite|improve this question







      New contributor




      Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




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      Paris is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 10 hours ago









      ParisParis

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          1 Answer
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          $begingroup$

          No, it is not.



          If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).



          If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
            $endgroup$
            – Paris
            8 hours ago








          • 2




            $begingroup$
            @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
            $endgroup$
            – Mike Shulman
            5 hours ago











          Your Answer





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          $begingroup$

          No, it is not.



          If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).



          If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
            $endgroup$
            – Paris
            8 hours ago








          • 2




            $begingroup$
            @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
            $endgroup$
            – Mike Shulman
            5 hours ago
















          7












          $begingroup$

          No, it is not.



          If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).



          If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
            $endgroup$
            – Paris
            8 hours ago








          • 2




            $begingroup$
            @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
            $endgroup$
            – Mike Shulman
            5 hours ago














          7












          7








          7





          $begingroup$

          No, it is not.



          If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).



          If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.






          share|cite|improve this answer









          $endgroup$



          No, it is not.



          If what you mean by $rm colim_n$ is the actual colimit of $F$ as a diagram of shape $Delta$, then this colimit is isomorphic to the coequalizer of the two maps $F([1]) rightrightarrows F([0])$. Coequalizers rarely preserve fibrancy, and with a little thought we can think of a counterexample that extends to a simplicial object. Let $M$ be a model category in which coproducts of fibrant objects are fibrant (e.g. simplicial sets), and let $G:Dto M$ be any diagram of fibrant objects whose colimit is not fibrant (such as a cospan $X leftarrow Delta[0] to Y$ in simplicial sets for almost any Kan complexes $X$ and $Y$). Let $F$ be the simplicial bar construction of $G$, with $F([n]) = coprod_{d_0 to cdots to d_n} G(d_0)$. Then each $F([n])$ is fibrant by our assumptions on $M$ and $G$, but the colimit of $F$ is the coequalizer of $coprod_{d_0to d_1} G(d_0) rightrightarrows coprod_d G(d)$, which is just the colimit of $G$ (it is the usual computation of colimits in terms of coproducts and coequalizers).



          If instead what you mean by $rm colim_n$ is actually the geometric realization, then there is an even easier counterexample. Let $M$ be simplicial sets, let $X$ be a simplicial set that is not a Kan complex, and let $F([n]) = X_n$ regarded as a discrete simplicial set. Discrete simplicial sets are Kan complexes, so each $F([n])$ is fibrant, but the geometric realization of $F$ is just $X$ itself, which is not fibrant.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Mike ShulmanMike Shulman

          36.8k483226




          36.8k483226












          • $begingroup$
            I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
            $endgroup$
            – Paris
            8 hours ago








          • 2




            $begingroup$
            @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
            $endgroup$
            – Mike Shulman
            5 hours ago


















          • $begingroup$
            I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
            $endgroup$
            – Paris
            8 hours ago








          • 2




            $begingroup$
            @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
            $endgroup$
            – Mike Shulman
            5 hours ago
















          $begingroup$
          I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
          $endgroup$
          – Paris
          8 hours ago






          $begingroup$
          I see, do you think it is still false if we impose that the two maps $F([1]) rightrightarrows F([0])$ are fibrations ?
          $endgroup$
          – Paris
          8 hours ago






          2




          2




          $begingroup$
          @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
          $endgroup$
          – Mike Shulman
          5 hours ago




          $begingroup$
          @Paris Yes, probably. Colimits rarely preserve fibrant objects and fibrations.
          $endgroup$
          – Mike Shulman
          5 hours ago










          Paris is a new contributor. Be nice, and check out our Code of Conduct.










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          Paris is a new contributor. Be nice, and check out our Code of Conduct.












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