In my work I've run into the following situation. In a model category, I have two reflexive coequalizers $A_i \stackrel{\to}{\to} B_i \to C_i$ and a map of diagrams which is levelwise a weak equivalence (i.e. $A_1\to A_2$ and $B_1\to B_2$ are weak equivalences). I need to conclude $C_1\to C_2$ is a weak equivalence. I'd love it if this were true all the time, but it probably isn't. Are there any standard axioms on a model category which let me conclude this is true? For example, left properness? I can't assume any of the objects are either cofibrant or fibrant, and the levelwise weak equivalences won't be fibrations or cofibrations.

Note that reflexive coequalizers often come up when one studies model categories because they are important for building objects of interest in categories of algebras over a monad (and to prove such categories are cocomplete). I've googled around quite a bit and can't find anything saying reflexive coequalizers preserve weak equivalences, so I'm pretty sure it's not true in general (and probably you can find a counterexample just in $Ch(R)$ via $A\otimes_R B$ or doing something similar to this) but I would really love to hear expert opinions on extra hypotheses which guarantee this.


1 Answer 1


Disclaimer: I am not an expert on model categories. $\newcommand{\MM}{\mathcal{M}} \newcommand{\pair}{\mathsf{P}} \newcommand{\dom}{\operatorname{dom}} \newcommand{\codom}{\operatorname{codom}} \newcommand{\colim}{\operatorname*{colim}} \newcommand{\hocolim}{\operatorname*{hocolim}} \newcommand{\id}{\mathrm{id}} \newcommand{\op}{\mathrm{op}} \newcommand{\Map}{\operatorname{Map}}$

Assume $\MM$ is a model category which verifies the condition — as required by the question — that reflexive coequalizers preserve weak equivalences. Then $\MM$ must be homotopically discrete in the sense that, for any two objects $X$ and $Y$ of $\MM$, the derived space of maps from $X$ to $Y$ is homotopically discrete. Here is a summary of the steps in a proof of that assertion.

  1. Any colimit can be written as a reflexive coequalizer.

  2. Then the condition required in the question entails that all colimits of objectwise cofibrant diagrams in $\MM$ are homotopy invariant.

  3. Therefore, under good technical conditions, colimits must actually be homotopy colimits. This will hold in particular for coequalizers of cofibrant objects in $\MM$.

  4. Applying to coequalizers of constant diagrams, we then show that $S^1\otimes X\to X$ (say, if $\MM$ is a simplicial model category) is a weak equivalence when $X$ is cofibrant.

  5. Finally, it follows that for any cofibrant $X$ and fibrant $Y$ in $\MM$, the space of morphisms $X\to Y$ is homotopically discrete.


For precision, let us first fix some terminology and notation: the category $\pair=(1\rightrightarrows 0)$ is a category with two objects, and generated by two parallel arrows from one to the other. A reflexive pair in a category $C$ is a diagram $\pair\to C$ such that the corresponding two parallel arrows in $C$ happen to admit a common section. A reflexive coequalizer is then simply the coequalizer of a reflexive pair.

The assumption on the model category $\MM$ described in the question asserts the homotopy invariance of reflexive coequalizers: if $F,G:\pair\to\MM$ are two reflexive pairs in $\MM$ and $\alpha:F\to G$ is a natural transformation which is objectwise a weak equivalence, then the induced map on the coequalizers $\colim F\to\colim G$ is also a weak equivalence. In this answer, we will assume this hypothesis holds for the model category $\MM$.

Step 1

Importantly, observe that any colimit in a cocomplete category $C$ can be functorially written as a reflexive coequalizer in the usual manner. If $F:I\to C$ is a small diagram in $C$ then its colimit is exactly the coequalizer of two parallel arrows $$ D_F \; : \qquad \coprod_{f\in I_1}F(\dom f)\rightrightarrows\coprod_{x\in I_0}F(x) $$ where $I_0$ and $I_1$ are the sets of objects and morphisms of $I$, respectively. Observe that this coequalizer is reflexive, i.e. there exists a common section to the two parallel arrows above. In fact, for any functor $F:I\to C$ one can construct a full simplicial object in $C$ such that the reflexive pair $D_F$ above is recovered as the image of the two maps $[0]\rightrightarrows[1]$ in $\Delta$.

Step 2

Note that the above reflexive pair $D_F$ is functorial in the diagram $F$. Therefore, if $\MM$ is a model category verifying the desired condition of homotopy invariance of reflexive coequalizers, then it will verify a more stringent property. Namely, small colimits of objectwise cofibrant diagrams are homotopy invariant in $\MM$: if $F,G: I\to\MM$ are two small diagrams in $\MM$ with cofibrant values, and $\alpha:F\to G$ is a natural transformation which is objectwise a weak equivalence, then $\alpha$ induces a weak equivalence on colimits $\colim F\to\colim G$.

Remark: We are assuming that $F$ and $G$ take cofibrant values in $M$ so that any coproduct of the maps $\alpha_x$ for $x\in I$ are also weak equivalences. For a general model category, we can only say that the coproduct of weak equivalences between cofibrant objects is a weak equivalence (see propositions 13.1.2 and 17.9.1 in Hirschhorn's book). Thus, the map induced by $\alpha$ on the reflexive pairs $D_F\to D_G$ is again a weak equivalence if $F$ and $G$ are objectwise cofibrant. Obviously, we can drop this restriction if coproducts in $\MM$ preserve weak equivalences.

Step 3

In summary, the requirement in the question entails that colimits of objectwise cofibrant diagrams in $\MM$ are homotopy invariant. Consequently, as long as the projective model structure exists on the category of functors $I\to\MM$, the colimit of any objectwise cofibrant diagram $F:I\to\MM$ is weakly equivalent to its homotopy colimit via the canonical map $\hocolim F\to\colim F$. This holds because the homotopy colimit of $F$ is actually the colimit of a cofibrant replacement of $F$ in the projective model structure on the category of functors $I\to\MM$.

It is well known that the projective model structure exists on the category of functors $I\to\MM$, for any small category $I$, whenever $\MM$ is cofibrantly generated. Regardless, the projective model structure always exists for the indexing category $I=\pair$, for any model category $\MM$. This is described in section 10 of Dwyer–Spalinski's Homotopy theories and model categories: specifically, see subsection 10.13 of that article.

Remark: Alternatively, even in the absence of the projective model structure, we can apply the formalism of the book of Dwyer–Hirschhorn–Kan–Smith titled Homotopy limit functors on model categories and homotopical categories. For any small indexing category $I$, given that colimits of objectwise cofibrant diagrams $I\to\MM$ are homotopy invariant, they must actually be homotopy colimits in the sense of that book. The formalism in that book is somewhat overkill, and we can probably instead use other similar frameworks for derived functors which encompass homotopy colimits in any model category: an example is this manuscript by Chacholski and Scherer.

Edit: Alternative to steps 2 and 3

The preceding step 3 has the advantage of being fairly conceptual, not depending directly on reflexive coequalizers (only on the conclusion from step 2). Notwithstanding, our focus on reflexive coequalizers allows for a simple approach which bypasses the steps 2 and 3 above, and does not require the existence of the projective model structure. For completeness, I explain it here.

In case $\MM$ is a simplicial model category, the homotopy colimit of $F:I\to\MM$ can be defined via a bar construction, giving $\hocolim F$ as the coequalizer of a reflexive pair $B_F:\pair\to\MM$ (see chapter 18 of Hirschorn's book, for example): $$ B_F \; : \qquad \coprod_{f\in I_1} B((\codom f) \downarrow I)^\op \otimes F(\dom f) \rightrightarrows \coprod_{x\in I_0} B(x\downarrow I)^\op \otimes F(x) $$ This diagram $B_F$ admits a natural projection to the previous reflexive pair $D_F$ whose coequalizer is $\colim F$. If $F$ takes cofibrant values, this projection is objectwise a weak equivalence, so we conclude that the natural map $$ \hocolim F=\operatorname{\colim_\pair} B_F \overset{\sim}{\longrightarrow}\operatorname{\colim_\pair} D_F=\colim F $$ is a weak equivalence.

When $\MM$ is an arbitrary model category (not simplicial), one should be able to give a similar argument by applying the definition of homotopy colimits in terms of framings of model categories, as described at the end of Hirschhorn's book.

Step 4

For simplicity, I will assume $\MM$ is a simplicial model category for the remaining steps. Nevertheless, we can get away with any model category $\MM$ (with no extra conditions) by using homotopy function complexes and framings in $\MM$ (as in Hirschhorn's book), or by working in the quasi-category associated with $\MM$.

Consider the coequalizer of the constant diagram $c_X:\pair\to\MM$ equal to a cofibrant object $X$ of $\MM$. Then $\colim c_X = X$. On the other hand, in a simplicial model category, the homotopy colimit of the constant functor $c_X$ is weakly equivalent to the tensor product of the classifying space of the index category with the (cofibrant) object $X$: $$ \operatorname{\hocolim_\pair} c_X = (\operatorname{\hocolim_\pair} 1)\otimes X\simeq B\pair\otimes X\simeq S^1\otimes X $$ where the first identity is a consequence of the definition of homotopy colimits in a simplicial model category as bar constructions (see the expression for $B_F$ above). Moreover, it also follows readily from this description that the map from the homotopy colimit to the actual colimit is induced by the unique map $!:S^1\to 1$: $$ \operatorname{\hocolim_\pair} X\simeq S^1\otimes X \overset{!\otimes\id_X}{\longrightarrow} 1\otimes X =X=\operatorname{\colim_\pair} X $$ As described in step 3, this map must be a weak equivalence under the hypothesis from the question.

Step 5

Now we easily conclude that $\MM$ is homotopy discrete. Let $X$ be a cofibrant object of $\MM$ and $Y$ a fibrant object of $\MM$. Since the above map $!\otimes\id_X:S^1\otimes X\to 1\otimes X=X$ is a weak equivalence, then the induced map $$ !^\ast: \Map(1,\MM(X,Y)) = \MM(X,Y) \overset{(!\otimes\id_X)^\ast}{\longrightarrow} \MM(S^1\otimes X,Y) = \Map(S^1,\MM(X,Y)) $$ is also a weak equivalence. Here $\MM(X,Y)$ denotes the simplicial set of morphisms from $X$ to $Y$ coming from the simplicial enrichment of $\MM$. Note that the preceding map $!^\ast$ has a left inverse: $$ i^\ast:\Map(S^1,\MM(X,Y)) \longrightarrow \Map(1,\MM(X,Y)) = \MM(X,Y) $$ for any choice of basepoint $i:1\to S^1$. Then $i^\ast$ is also a weak equivalence. The axioms for a simplicial model category imply that $\MM(X,Y)$ is a Kan complex, hence:

  • $i^\ast$ is a Kan fibration, and

  • the geometric realization of the fibre of the map $i^\ast$ at any $f\in\MM(X,Y)_0$ is equivalent to the space of based loops $\Omega_f\lvert\MM(X,Y)\rvert$.

Since $i^\ast$ is a weak equivalence and a fibration, its fibres must be weakly contractible, that is $\Omega_f\lvert\MM(X,Y)\rvert$ is contractible for any $f\in\MM(X,Y)_0$. In conclusion, $\MM(X,Y)$ has all its components weakly contractible, i.e. is homotopy discrete.

Concluding remarks

A priori, I would not expect to find any practical conditions ensuring that some useful class of reflexive coequalizers (or other colimits) in a model category are homotopy invariant, short of assuming some cofibrancy conditions on the diagrams. This belief is bolstered by the arguments above, and more broadly by the general ideology of model categories. I would certainly be very interested in learning of results or examples invalidating that view.

Moreover, in the specific case of algebras for monads, it may be a good idea to think about simplicial objects and geometric realizations instead of reflexive coequalizers. In particular, it is important to note that the usual reflexive coequalizer constructed from an algebra for a monad actually extends to an augmented simplicial object.

  • $\begingroup$ Hi Ricardo. Thanks for you answer. I need some time to process this, and unfortunately things are crazy right now in my personal life. So I'll read it when I can and hopefully be able to give meaningful feedback. For now it seems I have to prove what I want another way, but I already have an idea for that. So thanks! $\endgroup$ Mar 4, 2013 at 23:21
  • $\begingroup$ @David: You are most welcome! Please let me know if you find any issues with my answer. $\endgroup$ Mar 5, 2013 at 3:21

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