# Products in the homotopy category of a model category

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?

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

• 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.
– asdq
Commented Jul 25, 2019 at 15:39