The ordinary notions of limit and colimit are universal solutions to a problem, specifically, finding terminal/initial objects in slice/coslice categories. In the context of homotopy right Kan extensions (it's not hard to show that the theory of homotopy limits reduces to this case (the same holds for left Kan extensions/colimits)), we say that given a functor $f:C\to C'$ between small categories, and a functor $F:C\to A$ where $A$ is a cofibrantly-generated model category, that a natural transformation of functors $\alpha: H\to Ran_f F$ exhibits $H$ as the homotopy right Kan extension of $F$ if there exists an injectively fibrant replacement $G$ of $F$ such that the composite $H\to Ran_fF \to Ran_fG$ is a weak equivalence.

When $A$ is combinatorial, we can also simply define a homotopy right Kan extension functor along $f$ to be $Ran_f (Q (-))$, where $Q$ is a functorial injectively fibrant replacement functor.

This is easy enough to define, but why is this the definition? Why would we want to take fibrant/cofibrant replacements and consider their ordinary Kan extensions/limits/etc? I suspect that it has to do with the fact that homotopy is an honest equivalence relation on arrows from a cofibrant object into a fibrant object, but I would appreciate an actual explanation.

  • $\begingroup$ I can't help but wonder with all this topology you have been doing... will you make it to the midwest this fall? $\endgroup$ Sep 8, 2010 at 12:12
  • $\begingroup$ I got to the midwest on Sunday, and classes started yesterday, so don't worry so much! $\endgroup$ Sep 8, 2010 at 12:25
  • $\begingroup$ Also, I don't understand what you're getting at. $\endgroup$ Sep 8, 2010 at 12:26
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    $\begingroup$ Sean is referring to the Midwest Topology Seminar. $\endgroup$ Sep 8, 2010 at 12:28
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    $\begingroup$ Harry, have you looked at Mike Shulman's paper "Homotopy limits and colimits and enriched homotopy theory"? In it, Shulman works out an enriched version of the homotopical machinery of Dwyer, Hirschhorn, Kan, and Smith; some things actually simplify in the enriched setting, since you can work explicitly with the spatial enrichment that is "secretly" there. Anyway, left and right derived functors from $C$ to $Ho(D)$ are right and left Kan extensions (respectively) along $C\to Ho(C)$. $\endgroup$ Sep 8, 2010 at 18:42

3 Answers 3


There is a notion of homotopical Kan extension defined for "homotopical categories" (cats with a class of weak equivalences satisfying the 2-out-of-6 property).  I cannot go into much detail without making a bazillion definitions, but the reference is Dwyer, Kan, Hirschorn, and Smith's "Homotopy Limit Functors on Model Categories and Homotopical Categories".  Homotopical kan extensions and homotopy limit functors are defined, and model categories are demonstrated to be homotopically complete and cocomplete, and of course, fibrant and cofibrant replacement play a pretty big role in the proof.  I think of the book as MacLane's chapter on Kan extensions generalized to the homotopical case.


Here is a motivation for hocolims, for concreteness look at pushouts of topological spaces:

The usual pushout functor $colim: Top^P \rightarrow Top$, where $P:=\bullet \leftarrow \bullet \rightarrow \bullet$ is the diagram of a pushout datum, does not respect weak equivalences: Take for example the diagrams $pt \leftarrow S^1 \rightarrow pt$ and $D^2 \leftarrow S^1 \rightarrow D^2$ in $Top$. Since $D^2$ (the 2-dimensional disk) is contractible, there is a map between the two diagrams consisting of weak equivalences, i.e. a weak equivalence in $Top^P$. But if we apply colim to both diagrams we get non-equivalent objects in $Top$, namely $pt$ in the first case and $S^2$ in the second.

Thus the pushout functor $colim: Top^P \rightarrow Top$ does not take weak equivalences to weak equivalences and we can not invoke the universal property of $Ho(Top^P)$ to get a functor $Ho(colim):Ho(Top^P) \rightarrow Ho(Top)$ making the square commute which has $colim$ on top, $Ho(colim)$ at the bottom and the projections to the homotopy categories at the left and right. The best we can do is to find the terminal functor "$hocolim$" which allows filling this square with a natural transformation - which is exactly the Kan extension property you cited.

For model categories (but not only there, e.g. also Baues cofibration categories) you can define that functor via the top level, i.e. going $Top^P \rightarrow Top$, in a particularly neat way. Back to the example: Topologists noticed that the colim functor, when restricted to pushout data $A \leftarrow B \rightarrow C$ in $Top$ with $B$ cofibrant and the arrows cofibrations, does preserve weak equivalences. E.g. building the pushout of two inclusions of $S^1$ into contractible spaces such that there is "enough space" around the image of $S^1$ inside those spaces, you will always get something equivalent to $S^2$ - try it out! So the recipe for computing the $hocolim$ is first replacing your diagram by one with these properties (cofibrant replacement - this is an endofunctor which preserves weak equivalences) and then apply $colim$ - this does now preserve weak equivalences and thus descends to a functor between the homotopy categories.

So the intuition about $hocolim$ - which is good in great generality - is that it is the best approximation to $colim$ which preserves weak equivalences ("is homotopy invariant"); the cofibrant replacement construction stems from the fact that this is a class of objects where $colim$ is already homotopy invariant.

The story for homotopy limits is of course dual, instructive examples are homotopy fibers and homotopy fixed point objects.

Edit: Here is how you see that $colim \circ Q$ is $hocolim$ - assuming that we have a cofibrant replacement functor $Q$ on $Top^P$: you can simply check the universal property. So let $F:Ho(Top^P) \rightarrow Ho(Top)$ be a functor and $\tau:F \circ Ho_{Top^P} \rightarrow Ho_{Top} \circ colim$ a natural transformation. Here the functors $Ho_*$ are the projections to the homotopy categories which leave objects unchanged and map morphisms to their homotopy classes - I will omit them from the notation from now on; so we consider a natural transformation $\tau: F \rightarrow colim$ and have to show that it factors through $colim \circ Q$.

The cofibrant replacement functor $Q$ comes with a natural weak equivalence $Q \rightarrow id$. Composing with $F$ gives a natural isomorphism $F \circ Q \rightarrow F$. Now for each pushout datum $D \in Top^P$ we have the chain

$$F(D) \leftarrow F(QD) \rightarrow colim(QD) \rightarrow colim(D)$$

where the middle arrow is $\tau_{QD}$ (the natural transformation $\tau$ at the object $QD$) and the the outer two arrows arise by applying $F,colim$ respectively to $QD \rightarrow D$. The left arrow can be gone backwards because it is an isomorphism. The whole way from left to right is then equal to $\tau_D$ because of the naturality of $\tau$ (flip the outer arrows downwards, fill in $\tau_D$ below and you got the naturality square). This shows that each $\tau_D$ factors through $colim \circ Q$. To see that this factorization is natural in $D$, observe that $QD \rightarrow D$ and $\tau_{Q-}$ are natural in $D$.

Edit2: While Harry generously granted me a check mark after the above, Tom Goodwillie is of course right that a statement about uniqueness is in order. Here is why the above factorization is unique at the level of homotopy categories: Given any factorization $\tau=i \circ \tau'$ of our given $\tau$, for cofibrant $D$ it will factorize as

$$\tau_D=i_D \circ \tau_D':F(D) \rightarrow colim(QD) \rightarrow colim(D)$$

where the second arrow $i_D$ is an isomorphism, because it is $colim$ of a weak equivalence between cofibrant objects. So the first factor must be $\tau_D'=i^{-1}_D \circ \tau_D$, there is no choice here. For general $D$ we can express $\tau_D'$ as

$$F(D) \leftarrow F(QD) \rightarrow colim(QQD) \rightarrow colim(QD)$$ by walking around the naturality square for $\tau'$ for the morphism $QD \rightarrow D$. So we also have no choice for non-cofibrant $D$.

Edit3: As spotted by Tom Goodwillie I implicitly (and unconsciously!) used here that $i_{QD}=Q(i_D)$ - see the comments. You can prove this by drawing all available naturality diagrams with $QQQD$ at the left upper corner, seeing from those first that $QQ(i_D)=Q(i_{QD})=i_{QQD}$ and that hence $i_{QD} \circ QQ(i_D) = Q(i_D) \circ i_{QQD} = Q(i_D) \circ Q(i_{QD}) = Q(i_D) \circ QQ(i_D)$. Now one can cancel the isomorphism $QQ(i_D)$ on both sides.

Now I switch the notation back to mentioning the $Ho_*$: This showed that $Ho(colim \circ Q)$ is the Kan extension of $Ho_{Top} \circ colim$ along $Ho_{Top^P}$. That is already good to know, but it was just level zero of showing that $colim \circ Q:Top^P \rightarrow Top$ is the homotopy terminal homotopy invariant functor. To proceed you have to produce a simplicial set; e.g. do your localisation by passing to some model of an $(\infty, 1)$-category (e.g. hammock localisation, coherent nerve, ...), or use framings, or consider the nerve of some category of functors into $colim \circ Q$ and then show that a certain space is contractible. This should be done at a place where you can draw pictures, unlike in a MathOverflow answer.

  • $\begingroup$ Nice answer. I like the situation in simplicially enriched categories which are complete or cocomplete enough. There you can actually read off a universal homotopy coherent cone or cocone so as to visualise the universal property. The (co)simplicial replacement business of Bousfield and Kan is also a good way to gain an intuition of the holim or hocolim construction. Of course, this works best with weak equivalences being homotopy equivalences and so your point about fibrant and cofibrant objects is important. $\endgroup$
    – Tim Porter
    Sep 8, 2010 at 15:26
  • $\begingroup$ Peter, how does one prove that $hocolim:=colim(Q(-))$ is homotopically terminal in the "homotopical category of homotopical functors over colim"? $\endgroup$ Sep 8, 2010 at 18:13
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    $\begingroup$ To show that colimQ is homotopically terminal you must establish uniqueness (in an appropriate sense) as well as existence of the map from FD. But this can be done, too. $\endgroup$ Sep 9, 2010 at 1:08
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    $\begingroup$ I mean, of the map from F to colimQ (in the category that you get from the category of all homotopy-invariant functors over F by inverting the weak equivalences). $\endgroup$ Sep 9, 2010 at 1:55
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    $\begingroup$ Peter, I think that even to get the homotopy-category statement you need one more step: The two available maps $QQD\to QD$ are not a priori equal, even though are both equivalences. To see that they are equal you have to use their compositions with the three maps $QQQD\to QQD$. $\endgroup$ Sep 9, 2010 at 16:05

This could have been a comment to Peter's answer, except that it's going to be long. (Maybe I should be making it a new question.)

Suppose that $Ho(\mathcal C)$ is obtained from $\mathcal C$ by (universally) inverting a class $\mathcal W$ of morphisms. Let's call an object of $\mathcal C$ homotopically terminal if in $Ho(\mathcal C)$ it is terminal.

I did not assume anything about $\mathcal W$ (except that this localization existed). In practice we usually have more to work with, like a model structure or at least the 2 out of 6 condition. Given enough structure or information, there are ways of defining a simplicial set for each pair of objects such that this serves as a "function space", in particular such that its set of components is the set of morphisms in $Ho(\mathcal C)$. I'm not a master of the known sufficient conditions for making these function spaces -- I suppose that there are several overlapping approaches.

QUESTION: Can we say that (in any or all of these approaches) an object $X$ is homotopically terminal in the sense I gave above if and only if for every object $Y$ the function space $Hom(Y,X)$ is weakly equivalent to a point? I hope so.

Here is a general, relatively simple, approach to the idea of derived functor, really just a distillation of standard ideas as I understand them, but trying to steer away from extraneous optional technicalities:

Suppose that in $\mathcal C$ and in $\mathcal D$ there are classes of morphisms $\mathcal U$ and $\mathcal V$ respectively. Let $\mathcal F$ be the category of functors $\mathcal C\to \mathcal D$. In $\mathcal F$ let $h\mathcal F$ be the full subcategory of "homotopy-invariant" functors -- those which take $\mathcal U$ into $\mathcal V$. By a left derived functor of $F\in \mathcal F$ I mean the following: Form the category $hF$ of all homotopy-invariant functors over $F$. An object is a pair $(G,G\to F)$ where $G\in h\mathcal F$. A morphism is a map $G_1\to G_2$ compatible with the maps to $F$. Define the class of maps $\mathcal W$ in $hF$ by requiring that for every $X\in \mathcal C$ the map $G_1(X)\to G_2(X)$ belongs to $\mathcal V$. By a left derived funtor of $F$ let us mean a homotopically terminal object of the category $hF$ (with respect to this class $\mathcal W$).

The standard way to achieve such a thing is to use a "replacement functor" $Q:\mathcal C\to \mathcal C$. You need a map $Q\to 1$, and you need three things to be true: (1) $QX\to X$ always belongs to $\mathcal U$. (2) $Q$ takes $\mathcal U$ into $\mathcal U$. (3) As applied to maps between objects that are in the image of $Q$, $F$ takes $\mathcal U$ into $\mathcal V$. I believe that that's all you need to show that $F\circ Q$ is homotopically terminal in the sense I defined. The proof is basically contained in what Peter wrote, including the edits prompted by my comments.

You don't even need the 2 out of 3 condition for $\mathcal U$ or $\mathcal V$ to get this.


  • $\begingroup$ This is great (if it works)! $\endgroup$ Sep 10, 2010 at 17:30
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    $\begingroup$ The answer to your question is no. There is an essentially unique way to extract an "$\infty$-category" from a relative category; in fact, Barwick--Kan defined a model structure on RelCat which is Quillen-equivalent to all the other models of "$\infty$-categories". So, your hope boils down to the claim that if $C$ is an $\infty$-category and $c \in C$ is an object which becomes terminal in $ho(C)$, then it is (homotopically) terminal in $C$ itself. Any nontrivial connected $A_\infty$-space $X$ gives rise to a counterexample, namely $\mathrm{pt}/\!/X$. $\endgroup$ Jun 4, 2015 at 23:07
  • $\begingroup$ Err, I suppose I'm using "(homotopically) terminal" to mean "the spaces of maps to it are all contractible", which goes exactly against your definition. Sorry about that, I hope it's still clear enough what I meant. $\endgroup$ Jun 4, 2015 at 23:45

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