Let $F: \mathcal{C} \to \mathcal{D}$ be a left Quillen functor between model categories. In Definition 2.16 of Goerss–Schemmerhorn - Model Categories and Simplicial Methods, the left derived functor $LF: \mathcal{C} \to \operatorname{Ho}(\mathcal{D})$ is defined by $X \mapsto F(P)$. However, the article does not define the functor on maps.

In practice, this is often done under the assumption that $\mathcal{C}$ has functorial factorisation (or at least functorial cofibrant replacement), but Goerss–Schemmerhorn seem to imply that this assumption is not needed. Could someone help me understand a construction that works in general, without requiring functorial factorisation?

My best guess is to fix an arbitrary cofibrant replacement $P_X \to X$ for each $X \in \mathcal{C}$, and then try to construct a well-defined map $F(P_X) \to F(P_Y)$ in the homotopy category of $\mathcal{D}$, given a morphism $f:X\to Y$. We get morphisms $$ F(P_X) \rightarrow F(X) \overset{f}{\to} F(Y) \leftarrow F(P_Y). $$ If the right-hand arrow were a weak equivalence, then we would be done, since it would be invertible in the homotopy category. This is the case if $Y$ is cofibrant, since left Quillen functors preserve weak equivalences between cofibrant objects. They do not preserve weak equivalences in general though, so the approach doesn't seem to work.

  • 1
    $\begingroup$ Of course. Have a look at Quillen’s book. $\endgroup$ Jul 11, 2022 at 19:31
  • $\begingroup$ @FernandoMuro would that be "Homotopical Algebra"? I have looked at it but I'm struggling to find what I'm looking for. Is there any chance you could help me find the relevant part of the book? $\endgroup$ Jul 11, 2022 at 20:39
  • $\begingroup$ You don't even need a model category. $\endgroup$
    – user40276
    Jul 12, 2022 at 19:10

1 Answer 1


This depends on whether one insists on derived functors landing in the original model category (as is the case with modern approaches of Hinich and Dwyer–Hirschhorn–Kan–Smith), or in its homotopy category (as is the case with the classical approaches of Quillen, Grothendieck, etc.).

In the latter case it is indeed possible to define the left derived functor of a left Quillen functor in the original setting of (closed) model categories of Quillen, i.e., without the existence of functorial factorizations and small (co)limits, added later by Kan.

Given a left Quillen functor $$F\colon C→D,$$ pick a cofibrant replacement (not necessarily functorial) $Q_X$ and an acyclic fibration $q_X\colon Q_X→X$ for every object $X∈C$. Now construct the left derived functor $$\def\ldf{{\bf L}}\def\Ho{{\rm Ho}} \ldf F\colon C→\Ho(D)$$ as follows. Send an object $X∈C$ to $Q_X∈\Ho(D)$. Given a morphism $f\colon X→Y$ in $C$, factor the composition $f∘q_X\colon Q_X→Y$ through the acyclic fibration $q_Y\colon Q_Y→Y$ obtaining a map $g\colon Q_X→Q_Y$. Send $f$ to the morphism $$F(g)$$ in $\Ho(D)$.

To see that this construction yields a functor, consider composable morphisms $f\colon X→Y$, $g\colon Y→Z$ together with their chosen lifts $Q_f\colon Q_X→Q_Y$, $Q_g\colon Q_Y→Q_Z$, as well as the lift of the composition $Q_{gf}\colon Q_X→Q_Z$. Now $Q_g Q_f\colon Q_X→Q_Z$ is another lift for $gf\colon X→Z$. Any two lifts of the same morphism through an acyclic fibration are left homotopic (see, for example, Proposition 1.2.5(iv) in Hovey's book). Thus, the two lifts map to the same morphism in $\Ho(D)$.

An elementary argument then shows that a different choice of $q$ and $Q$ produces the same morphism in $\Ho(D)$, and therefore the same functor $C→\Ho(D)$.

Added: This is essentially the construction given by Quillen in Proposition I.4.1 of Homotopical Algebra, except that Quillen's construction starts by defining a functor $C→\Ho(C_c)$, where $C_c$ denotes the full subcategory of $C$ on cofibrant objects. This functor is then composed with $F$, using Ken Brown's lemma.

  • $\begingroup$ I know I already accepted the answer, but would you mind expanding on what the elementary argument is that shows we have a functor? The only things I know about the homotopy category are the universal property and the fact that homotopic maps get identified. My understanding of homotopic maps is pretty flimsy though, so I'm not sure if it's obvious that different choices for F(g) are homotopic $\endgroup$ Jul 12, 2022 at 12:48
  • 1
    $\begingroup$ @SebastianMonnet: I added an argument for this. $\endgroup$ Jul 12, 2022 at 16:47
  • $\begingroup$ That's very helpful, thank you! $\endgroup$ Jul 12, 2022 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.