Given a model category $C$, a simplicial symmetric monoidal model category $D$ (in the sense of Goerss-Jardine) and a left Quillen functor $F:C\to D$, define $|F|:C\to\mathsf{sSet}$ to be the functor $$|F|=\mathsf{Map}_D(I,F(-))$$ where $I$ is the unit of the monoidal structure of $D$ which we suppose to be cofibrant.

Then, is there a link between $|F|$ and $|\mathbb LF|$, like $\mathbb L|F|\simeq |\mathbb L F|$ ? The problem is that $|F|$ may not verify the classical conditions to have a left derived functor, as we want fibrations on the right argument of the mapping space.

In my case $D=C(k)$ for $k$ a field of characteristic $0$, so every object is fibrant, which might help. My functor $F$ computes something which has the good homotopy type only for cofibrant objects, so I'd want $\mathsf{Map}_D(I,F(Q(-)))$ to yield a functor $\mathsf{Ho}(C)\to\mathsf{Ho}({\mathsf{sSet}})$ and I'd like it to be the total left derived functor of $|F|$.

  • $\begingroup$ The final object in D=C(k) is the zero chain complex, so *=0 and Map(*,F(-))=0. It would appear then that your question is trivial. $\endgroup$
    – Dmitri P.
    Commented Jul 16, 2019 at 0:58
  • $\begingroup$ Fixed, thanks. In my case this is $k$. The reference to Goerss Jardine isn't really correct now, but anyway, just take $D=C(k)$ :) $\endgroup$
    – elidiot
    Commented Jul 16, 2019 at 7:00

1 Answer 1


If I is cofibrant and all objects are fibrant, then Map(I,−) is automatically derived. Thus, precomposing with the cofibrant replacement functor Q suffices to derive both |F| and F.

We get L|F|=|QF| and |LF|=|QF|, so indeed there is a weak equivalence between L|F| and |LF|.

  • $\begingroup$ Thank you. Yet I'm wondering : as a functor from $C$ to $\mathsf{sSet}$, $|F|$ does not seem to preserve cofibrations or anything like this, so how do we know that it actually has a left derived functor ? $\endgroup$
    – elidiot
    Commented Jul 17, 2019 at 9:15
  • $\begingroup$ @elidiot: Preservation of cofibrations has nothing to do with the definition of left derived functors. In this case, |F| has a left derived functor because Q (with the natural weak equivalence Q→id) is a homotopy equivalence of relative categories and the composition |F|∘Q preserves weak equivalences. $\endgroup$
    – Dmitri P.
    Commented Jul 17, 2019 at 15:08
  • $\begingroup$ Ok I was confused, thanks. I have one last problem : why would $|F|\circ Q$ perserve weak equivalences ? $\endgroup$
    – elidiot
    Commented Jul 17, 2019 at 15:58
  • $\begingroup$ @elidiot: Map(I,−) preserves weak equivalences because I is cofibrant and all objects are fibrant. F∘Q preserves weak equivalences by Brown's lemma. Therefore, the composition of these functors, Map(I,F∘Q)=|F|∘Q preserves weak equivalences. $\endgroup$
    – Dmitri P.
    Commented Jul 17, 2019 at 16:06
  • $\begingroup$ Oh, what I didn't know is that Map(I,-) preserves weak equivalences. But I guess it is Brown's lemma again as it sends trivial fibrations between fibrant (=all) objects to equivalences (I mean, using only the axioms of a simplicial model category). Thank you very much for your help. $\endgroup$
    – elidiot
    Commented Jul 17, 2019 at 16:20

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