Why is the category of motives generated by varieties?

Question

I'm reading Ayoub's paper Motifs des varietes analytiques rigides, but I'm not quite familiar with motives. In this paper, he defines the category of motives to be $\mathbf{RigDM}^{\rm eff}_{\rm Nis}(k, \mathbb{Q}) := \mathbf{Ho}_{\rm Nis, \mathbb{B}^1}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$.

As I get it, we consider the site of all smooth rigid spaces over $k$, then consider all complexes of presheaves over this site. Just like the construction of derived category, now we invert all quasi-isomorphisms as Nisnevich sheaves, and invert all $\mathbb{B}^1-$homotopy, namely morphisms like $\mathbb{Q}(\mathbb{B}^1\times X)[i]\rightarrow \mathbb{Q}(X)[i]$.

What I'm interested in is the Theorem 2.5.35, it says that this category can be compactly generated by those $\mathbb{Q}(X^{\rm an})[i]$ with $X$ being proper smooth $k$-schemes.

As I follow the proof, the first step is the fact that the category $\mathbf{Ho}_{\rm Nis}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$(So hence $\mathbf{RigDM}^{\rm eff}_{\rm Nis}(k, \mathbb{Q})$) is compactly generated by those $\mathbb{Q}(X^{\rm an})[i]$ with $X$ being quasi-compact $k$-schemes. Then he uses methods of algebraic geometry to reduce $X$ to be those proper smooth ones.

However, I don't quite follow this first step. In the category of modules, since we allow taking arbitrary direct sums, from $\mathbb{Q}$ we get all free modules. And taking direct summands is also allowed, so we have projective modules. Then by projective resolution, we get all modules. And in the derived category, all complex is quasi-isomorphic to its cohomology, so we get the whole derived category.

I guess for sheaves, it also follows this process to generate the category of motives. As I understand, the sheaf $\mathbb{Q}(X)$ that $X$ represent is just something as a constant sheaf on $X$. However, I don't quite understand how this goes, and I don't even know if there are enough projective sheaves for this topology...

Answer 1

No, $\mathbb{Q}(X)$ is the presheaf "additively represented" by $X$; it is not constant.

Now I will express my understanding of this matter; I did not check that it fits with Ayoub's definitions and notation, sorry. I believe that the category $\mathbf{Ho}_{\rm Nis}(\mathbf{Compl}(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q})))$ is just a localization of the category $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$, and the localization functor respects coproducts. Thus it suffices to study $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$. Now, this category is generated (as its own localizing subcategory) by the images of $\mathbb{Q}(X)$ essentially by definition. Being more precise, this derived category is defined as the localization of $K(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ by the subcategory of objects orthogonal to all $\mathbb{Q}(X)$ (look at $O$ with only zero morphisms from all shifts of $\mathbb{Q}(X)$ into it). Thus there are no non-zero objects in $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ that are orthogonal to $\mathbb{Q}(X)$. Since objects of the type $\mathbb{Q}(X)$ are compact, it follows that they generated $D(\mathbf{PreShv}(\text{SmRig}_k, \mathbb{Q}))$ as its own localizing subcategory indeed (this is a well-known fact; see Proposition 8.4.1 of Neeman's "Triangulated categories").

Lastly, this "orthogonality argument" gives an easier (though a "less explicit") proof of the compact generation of $D(R)$ as well.