Consider the category $\operatorname{Set_\Delta}$ of simplicial sets, equipped with a model structure such that it becomes a cartesian closed model category (e.g. the Quillen or Joyal model structure). Let $S$ be any simplicial set. There is a Quillen adjunction $(S\times -, [S,-])$ between the product and internal hom functor, therefore one obtains an adjunction between the corresponding left and right derived functors on the homotopy category. I would like to conclude that $\mathbb R[S,-]$ gives exponential objects in the homotopy category, which is equivalent to proving that $\mathbb L(S\times -)$ computes products in the homotopy category.

Unfortunately, I seem unable to do so. According to the nlab this is supposed to be true, but I could not find any other reference where this is stated (let alone proved). Can anybody help me out?


1 Answer 1


I'm not very knowledgeable on model categories, so I don't know how this relates to your precise claim, however the claim the nLab makes (homotopy product = product in the homotopy category) is quite easy to prove (the reason I don't know how this relates to your claim is that you are considering $\mathbb L(S\times -)$, whereas the homotopy product is the right derived functor of the product functor $\mathbf{sSet}^2 \to \mathbf{sSet}$).

Let $C$ be a model category. Unless indicated otherwise, $\times$ denotes the usual product in $C$; $[-,-]$ denotes the hom-set in the homotopy category.

Let $X,Y$ be fibrant, then clearly the image of $X\times Y$ is the product of $X$ and $Y$ in the homotopy category : if $Z$ is any object, then one may cofibrant replace it and so assume that $Z$ is cofibrant, so $[Z, X\times Y]$ is isomorphic to homotopy classes of maps $Z\to X\times Y$ and therefore to pairs of homotopy classes of maps $Z\to X, Z\to Y$ (here we use the fact that we can choose a cylinder object for $Z$ uniformly, and any homotopy can be realized via this fixed cylinder object), and therefore to $[Z,X]\times [Z,Y]$.

Therefore, if $X,Y$ are any objects, then you fibrant replace them with $QX,QY$ and get that their homotopy product is $QX\times QY$ which also happens to be the product of $QX$ and $QY$ in the homotopy category, which is therefore the product of $X$ and $Y$ in the homotopy category (because $X\simeq QX, Y\simeq QY$ in the homotopy category obviously)

So $X\times_h Y = X\times_{\mathrm{Ho}(C)} Y$

  • $\begingroup$ Thank you for your answer. As for your first sentence, this is what I realized after I published my question, so there is perhaps more to it. But your post definitely answers the question in the way I posed it. $\endgroup$
    – asdq
    Commented Jul 25, 2019 at 15:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .