# Small fppf/syntomic/smooth sites?

I was reading the section about sites in the Stacks Project. Small sites are studied only in few cases, but I don't get why. As some users say, it's because of this.

If $f: X\to Y$ is a morphism of schemes, it's known it's not generally true that the small fppf inverse image functor $f^{-1}$ is exact, as opposed to $f_{small, etale}^{-1}$, $f_{small, Zariski}^{-1}$, $f_{small, crystalline}^{-1}$, and all the big inverse images.

Any explicit counterexample for $f^{-1}_{small, fppf}$, $f^{-1}_{small, smooth}$, $f^{-1}_{small, syntomic}$, etc. ?

Related - Leo Alonso's comment to Etale site is useful - examples of using the small fppf site?

It'd be nice to collect some examples here.

EDIT. Merlin's example says it all. $f^{-1}_{small, *}$ is not exact for $*\in\{smooth, syntomic,fppf, fpqc, Lisse-\acute{E}t, Flat-fppf\}$, so this is why Grothendieck switches to big sites, as Leo Alonso says in his comment (see first link).

Good point. Meanwhile I'll say a few things, which I plan to make into an answer at some point, though I suggest you wait for a bunch of examples to come from users all over.

To fix ideas, we consider the category of all smooth $S$-schemes with arbotrary $S$-morphisms between them, denoted $\mathcal{C}$, and endow it with the smooth topology. The resulting site will be the small smooth site for us $S_{\text{smooth}}$ (details omitted on checking this is indeed a site).

Given a morphism of schemes $f : S'\to S$, asking whether $f^{-1}$ is exact is really a question about $\mathcal{C}$ and not the topology $\mathcal{C}$ is endowed with. For $f^{-1}$ to be exact, the base-change functor has to commute with all finite limits, hence finite limits must, first, exist in $\mathcal{C}$. This is equivalent to asking that all finite products and equalizers exist in $\mathcal{C}$. Here we find a first obstacle: if you equalize two $S$-morphisms $a,b: U\to V$ in the category of $S$-schemes, for objects $U\to S$ and $V\to S$ in $\mathcal{C}$, the equalizer $E$ of $a,b$ in schemes over $S$ won't be $S$-smooth in general, ie. $\mathcal{C}$ doesn't contain equalizers, so we're screwed (strictly speaking, $\mathcal{C}$ may still contain equalizers, and they may not agree with equalizers in the category of schemes. You should check this is not the case).

For an explicit example, you may take a ring $R$, $S = \text{Spec}(R)$, $U = V = \mathbb{A}^1_S = \text{Spec}(R[t])$, and the maps:

$a : U\to V$ given by $t\mapsto 0$, and $b: U\to V$ given by $t\mapsto t^2$.

Their equalizer in the category of $S$-schemes is $E := \text{Spec}(R[t]/(t^2))\to S$, which is not $S$-smooth.

• Thank you. Your example doesn't seem to obstruct exactness of $f^{-1}_{syntomic}$, for instance
– user95222
Aug 8, 2016 at 5:23
• I must have seen a similar trick somewhere. Take the identity $f: \mathbb{A}^1_{\mathbf{Z}}\to\mathbb{A}^1_{\mathbf{Z}}$ and then the map $g$ induced by $t\mapsto t+p$ on $\mathbf{Z}[t]$. Their equalizer in $\mathbf{Z}$-schemes is $\mathbb{A}^1_{\mathbf{F}_p}$, surely not syntomic over $\text{Spec}(\mathbf{Z})$. So I guess small syn inv image along $\text{Spec}(\mathbf{F}_p)\to\text{Spec}(\mathbf{Z})$ is not exact.
– user87684
Aug 8, 2016 at 10:34
• And this is also a counterexample to exactness of $f^{-1}_{small, fppf}$ and $f^{-1}_{small, smooth}$. Great.