# Why is every object cofibrant in an excellent model category?

In Appendix A.3 of the book higher topos theory appears the notion of an excellent model category (see Definition A.3.2.16). The main feature of this notion is that when $\mathbf{S}$ is an excellent model category one has a model structure on the category of small $\mathbf{S}$-enriched categories which enjoys several good properties. Condition (A2) in the definition of an excellent model category states that every monomorphism is a cofibration and that the collection of cofibrations is closed under taking products. In remark A.3.2.17 which appears right after the definition it is said that condition (A2) implies that every object of $\mathbf{S}$ is cofibrant. This remark is reiterated (and relied upon) in the very nice paper http://arxiv.org/abs/1602.05313 of Tyler Lawson where it showed that the first four conditions in the definition of an excellent model category imply the fifth (which is essentially the only one which is hard to check in practice).

I am probably just missing something trivial here, but I just don't see why condition (A2) implies that every object is cofibrant. This would be the case, of course, if the map $\emptyset_{\mathbf{S}} \to X$ from the initial object would always be a monomorphism, but this is not true in general, not even for presentable categories (for example, in the category of rings, the map $\mathbb{Z} \to 0$ from the initial ring to the terminal ring is not a monomorphism).

My question is then: how does condition (A2) imply that every object is cofibrant?

• I bet that you should also require the tensor unit to be cofibrant. This us actually assumed in other parts of the book. Jun 24 '16 at 18:39
• I'm happy to assume the unit is cofibrant, but I still don't see how the result would follow. Jun 25 '16 at 16:09
• Yonatan, tensor the cofibration from the initial object to the unit object with the identity in X, which is another cofibration. Jun 25 '16 at 18:35
• If this argument worked it would prove that in every monoidal model category with a cofibrant unit every object is cofibrant. The thing is that the pushout-product axiom doesn't imply that tensoring two cofibrations is a cofibration. In particular, tensoring a cofibration with an object $X$ is only guaranteed to be a cofibration if you already know that $X$ is cofibrant. Jun 25 '16 at 19:12
• Sorry, I took product for tensor product in A2. Then either you also need A2 for tensor products or you rather ask that the final object is cofibrant. Jun 25 '16 at 20:18

I am now under the impression that it is simply not true that in an excellent model category every object is cofibrant. Let $\mathbf{S}$ be an excellent model category in which the monoidal structure is the Cartesian one (e.g., simplicial sets). For every $X \in \mathbf{S}$ we may consider the coslice category $\mathbf{S}_{X/}$ with its canonical model structure, in which all three classes are created by the projection $p:\mathbf{S}_{X/} \to \mathbf{S}$. We now note that $p$ creates limits and connected colimits (i.e., colimits indexed by connected diagrams), and in particular pushouts. It then follows that the Cartesian product on $\mathbf{S}_{X/}$ is compatible with the model structure. We now claim that $\mathbf{S}_{X/}$ is an excellent model category (with respect to the Cartesian monoidal structure). First note that $\mathbf{S}_{X/}$ is combinatorial. Property (A2) discussed above follows from the same property for $\mathbf{S}$ since $p:\mathbf{S}_{X/} \to \mathbf{S}$ is a right functor and hence preserves products and monomorphisms (while creating cofibrations by definition). Weak equivalences in $\mathbf{S}_{X/}$ are closed under filtered colimits since $p$ preserves filtered colimits and detects weak equivalences. Finally, the invertibility hypothesis follows from Lemma A.3.2.20 of HTT since we have a left Quillen functor $\mathbf{S} \to \mathbf{S}_{X/}$.
On the other hand, $\mathbf{S}_{X/}$ will typically not have all its objects cofibrant. Take for example $\mathbf{S}$ to be simplicial sets and $X = \Delta^0 \coprod \Delta^0$, in which case the object $X \to \Delta^0$ is not cofibrant in $\mathbf{S}_{X/}$.
Edit: Actually the monoidal product on $\mathbf{S}_{X/}$ cannot not be taken to be the Cartesian one, because the Cartesian product on $\mathbf{S}_{X/}$ doesn't preserve colimits in each variable separately, so this argument doesn't work as is. However, all the rest seems to still be correct, so that $\mathbf{S}_{X/}$ will be an excellent model category as soon as it has some compatible monoidal structure. In particular, $\mathbf{S}_{X/}$ satisfies (A2) but not every object is cofibrant in general.
Second Edit: Constructing monoidal structures on $\mathbf{S}_{X/}$ turned out to be quite of a headache. However, there is a natural left action of $\mathbf{S}$ on $\mathbf{S}_{X/}$ which is given by $$Z \otimes (X \to Y) := X \to X \coprod_{Z \times X} \left[Z \times Y\right]$$ and a natural right action given by $$(X \to Y) \otimes Z := X \to X \coprod_{X \times Z} \left[Y \times Z\right] .$$ Here the pushouts are performed with respect to the projections $Z \times X \to X$ and $X \times Z \to X$ respectively (these two actions are naturally isomorphic, but it is convenient to have them written on both sides). These actions preserve colimits in each variable separately, and it is not hard to verify that if the Cartesian product on $\mathbf{S}$ satisfies the pushout-product axiom then these actions satisfy the pushout-product axiom as well. We may hence think of $\mathbf{S}_{X/}$ as a bimodule model category over $\mathbf{S}$. In this case we can construct a new (non-Cartesian) monoidal model category, the "square-zero extension" $\mathbf{S}_{X/} \rtimes \mathbf{S}$, whose underlying model category is the product model category $\mathbf{S}_{X/} \times \mathbf{S}$, and such that $$(Z,X \to Y) \otimes (Z',X' \to Y') := (Z \times Z',\left[Z \otimes (X' \to Y')\right] \coprod \left[(X \to Y) \otimes Z'\right])$$ When $\mathbf{S}$ is an excellent Cartesian model category (e.g. simplicial sets) then $\mathbf{S}_{/X}$ satisfies all the axioms of an excellent model category except the existence of a monoidal structure (see above). This implies that the monoidal model category $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is actually an excellent model category. However, in general not every object in $\mathbf{S}_{X/} \rtimes \mathbf{S}$ is cofibrant, e.g. when $\mathbf{S}$ is simplicial sets and $X = \Delta^0 \coprod \Delta^0$.
The conclusion of this discussion is not that provocative. Since the condition that every object is cofibrant is required (or at least used) for setting up a model structure on $\mathbf{S}$-enriched categories (see Proposition A.3.2.4 of HTT), all this means is that one should simply add "every object is cofibrant" to the list of axioms defining an excellent model category.