# 2.2. 概形是 fppf 层 (仍然在翻译中)

Next, we will show if is a scheme, then is sheaf for the fppf topology.

We expand the definitions to see what this means. We see all we get flat locally of finite presentation and surjective. We also get exact sequence Then, sheaf for fppf topology means we getthen there exists unique arrow from to .

A special case of this is that, open cover and is equivalent to such that .

The big theorem we are working towards at this point is that, if is a scheme, then is a sheaf for the fppf site on all schemes, i.e. the fppf site on . In particular, if then is a sheaf on the fppf site on .

Since Zariski and etale covers are examples of fppf covers, the big theorem also says is a sheaf for the (big) etale and Zariski topologies. We will start prove the theorem.

Next, we check exactness on the middle. Suppose , then we have . In particular, we getThis concludes the exactness on the middle as well.

Exact on the left: Suppose we have with and injective. However, this implies as is injective, as desired.

Exact on the middle: Say such that for all , . By previous proposition, we knowHence factors through , which proves our claim.

Exactness on the left: We need to show injects into . Let be open affine cover and be fppf. Then be an open affine cover. Hence . We get the following diagramThus, to prove the exactness on the left, it is enough to show the bottom arrow is injective. Since is flat locally of finite presentation, we see is open. Thus are open in . Also, is surjective as its fppf, we seewhere the each are open. Since is affine, it is quasi-compact, we see we can take finite subcover . In particular, we see the mapis fppf becausewhere the inclusion is flat. Thus we see since and are both affine, so by assumption, we getwhich shows exactness on the left.

Exactness on the middle: Suppose is fppf with not necessarily affine. We will do step by step.

Step 1: we show we may assume affine. Let is affine open cover. Let . Then we getWe get injective since is big Zariski sheaf, the is injective since is fppf and we apply exactness on the left shown above. A diagram chase shows exactness in the middle as desired (if we can show the affine case).

The diagram chase is roughly as follows: start with the top middle bullet , we want to ask if there exists in that maps to :There is not much we can do at this point, thus we send to the bottom via and getWe want to show comes from . Thus we getHowever, note the middle row is exact by assumption, we indeed getViz, we have comes from and it remains to show comes from the injection arrow from . To that end, we note the left vertical line is exact, hence to show lives in the image of , we just need to show has the same image in . To show that, we map forward via the two different maps, and get and , i.e. we getwhere at the bottom, we must have and map to the same element because the middle row is exact and the image of and must equal the image of . Hence, this forces which forces to come from and hence shows indeed comes from as desired.

Step 2: we show we can assume is quasi-compact. We showed last time there exists open cover by quasi-compacts such that fppf. Consider the restriction mapOur goal is to show comes from .

Assume quasi-compact case. Since is fppd, we get the following sequenceis exact. Thus there exists unique mapping to . We claim is independent of . Indeed, consider the diagramwhere we used fppf maps and to get the fibered product. Now we apply to the whole diagram. First, we get the following injectionsNow let’s chase elements:However, since maps to and , we know and must map to the same thing in . However, is fppf cover, thus the two arrows are injective. Thus, we must have , hence we can denote this as . Moreover we have . This is exactly what we wanted, and hence this finishes step .

Step 3: finish the proof. We may assume fppf with quasi-compact and affine. Let be a finite affine cover. In particular, since the union is finite, we see is affine and hence is fppf.

Thus we getThe vertical arrows are injective since is fppf cover. The bottom row is exact because is fppf and both of them are affine and we are assuming the affine case holds. Hence the top row is exact. This concludes the proof.

Last time, we proved a big lemma says if is sheaf for big Zariski topology, then is sheaf for fppf topology iff for affine singleton covers we get the exact sequence.

The next step is to show we can move from affine to any .

Exact on the left: say we havesuch that . We want . Set-theoretically, we know as is surjection. Thus we just need to show this equality is scheme-theoretically. Since and as set maps, we see . Thus let’s define . Now take the diagram and restrict to with , we getIn particular, we get . However, is a sheaf as is affine, thus we see scheme-theoretically. However, this holds for all and hence we see scheme-theoretically globally.

Exact on the middle: say we havewith . We want to show the dash arrow exists, i.e. we want to show there exists . Let be the underlying topological space of any scheme . Then we see we get the same diagram for topological spacesHowever, in this case, exists because of the following claim.

Claim:is a coequalizer of topological spaces.

Suppose this claim holds, then exists topologically , and so we can talk about subschemes , . Then, we getwhere the existence of is by affine case. Moreover, we see because we can cover by affine open and restrict to open affine agrees with restricts to open affine by the uniqueness statement in the affine case. Since , the ’s glue to a map scheme-theoretically. Thus, if we can prove the above claim, we are done. We are going to prove it in small steps.

Next, we claim that for fppf ,:

 1 there exists natural surjection 2 open iff is open. 3 is coequalizer in category of topological spaces.

: Take with the same image . Then we see we getWhat we do next is to take fibered products on the right hand side (of the residue fields) and we getwhere we see the right hand side is and we just choose any point of this. This yields a point of mapping to .

: is open then since is continuous we get is open. Conversely, is fppf implies it is surjective, thus and hence its open as is open and is open map.

: we need to show the diagram is a coequalizer diagram. In this part, we will drop the bars, and just move to the category of topological spaces. What we want is that for any topological space, we wantIf exists, then it is unique as is surjective, i.e. , then choose such that , then .

Thus we just need to show if and , then . However, since with , this means . Earlier, we showed there is surjectionLet so . Then, we see we know by assumption, thus we see but by definition and .

At this point, we have defined as set map, and we need to show is continuous. Thus take be open, then we see is open iff is open in by part . However,where is continuous, hence is continuous as desired.