The Stacks project

OK, I understand your confusion. When we have a ring $A$ and an element $f \in A$, then the notation $A_f$ and $M_f$ always mean: $S^{-1}A$ and $S^{-1}M$ where $S$ is the multiplicative subset $S = \{1, f, f^2, f^3, \ldots\}$ of $A$ even when $f$ is nilpotent (in which case $A_f$ is the zero ring and $M_f$ is the zero module). Definition 107.22.1 really means for every $\mathfrak p$ there is an $f \in A$, $f \not \in \mathfrak p$ such that $M_f$ is a finite free $A_f$-module. But note that any module over the zero ring is both zero and free and finite and invertible (either by convention or because this follows from the definition of a free or finite or invertible module).

Or is it that in this definition, localization at a point $f \in$ A means that f is a non-nilpotent element and the set S = {$f^n$ : n = 0,1,...}? Thanks in advance!

I think this is a regular induction, not a descending induction.

In Definition 107.22.1 for a finite locally free A-module, it should be "for every p $\in$ Spec(A), $\exists$ an $f \in$ Spec(A)" instead of "f $\in$ A" because otherwise it does not make sense to localize A at f.

Oops, this is very confusing, because there is a mistake, namely, we should replace the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Ker}(d^{i + 1})$ by the condition that $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$. Thanks for pointing this out. I will fix this very soon.

Let me explain a bit more: for any choice of $E^i \subset X^i$ with $d^i(E^i) \subset D^{i + 1} + E^{i + 1}$ we see that setting $Y^i = D^i + E^i$ gives a subcomplex whose terms are in $\mathcal{B}$ and whose cohomology surjects onto the cohomology of $X^\bullet$. Clearly, if $d^i(E^i) = (D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i)$ then we see that the map on cohomology is also injective. So now start with some really large index $n$. Then we take $E^n$ equal to $0$. By descending induction we can choose $E^i$ for all $i$ with the desired property.

For example, for $i$ the largest integer such that $X^i$ is nonzero we can choose $E^i = 0$ because this will surject onto $(D^{i + 1} + E^{i + 1}) \cap \text{Im}(d^i) = 0$.

I fail to follow the proof of the claim in Lemma 0FCL. I don't think the injectivity of the induced map on cohomology is obvious.

The construction of E^i is also confusing to me. For, take i=n, which is the largest number that X^i is nonzero. What is E^n?

So I was wondering it might be a good idea to add more details...

The citation to WeilII here actually points to Weil I.

Unless this appears elsewhere, it might be useful to add that if $X$ is reduced then one can weaken statement (5) to '$\mathcal{F}$ is finite type and $\rho_{\mathcal{F}}$ is locally constant'. i.e. the requirement that $\mathcal{F}_x$ is free can be dropped.

(This appears as Exercise 13.7.K in the 17/11/2017 version of Vakil's notes The Rising Sea or Exercise 5.8 in Hartshorne.)

The affine case Lemma 00NX could also include the corresponding statement for finite type modules over reduced rings.

I believe part (1) is actually proven in 03PA. It may be helpful to include a link to that here.

Some typos:

"Thus if P holds for F1 and F2, then it hods for F"

"To finish the proof have to show that Hd(X,i∗I)=Hd(Z,I) ..."

"As G|U=KL⊗n|π−1(U) is invertible ..." (extra mathcal K here)

Singular cohomology is outside the ambit of the Stacks project... at least for now.

Of course! Thank you.

Minor typo: "On the same underlying category with have a second topology, namely the étale topology"

Is there any section on comparison between etale cohomology and singular cohomology? If not, are there any plans to add such a section?

Yes, it is because $h$ restricted to the open $X_U$ is given to us and $X$ is the union of the opens $X_U$.

Are the morphisms $h:X\to Y$ unique?

No problem!

Sorry, my bad. You are of course completely correct and you said it correctly the first time too. I will fix this when I next go through all the comments. Thanks!

I'm sorry, either I am missing something else, or I did not express myself clearly. Let us assume $F$ is infinite (as in the proof of the lemma, I agree with you that most people would apply this to the parenthetical comment). This is false: a finite union of proper subvector spaces is never a subvector space. Counterexample: If $U,W$ are proper subvector spaces of $V$, then $U\cup W$ is a subvector space as long as $U\subseteq W$ or $W\subseteq U$. This is true (and I think this is what was intended): a vector space cannot be written as a finite union of proper subvector spaces.

@#4595 Yes, and just two sentences earlier we reduced to the case where $F$ is infinite. I think most people would read the parenthetical remark as applying to that case.

In the proof of the lemma, the statement (in the second paragraph of the proof) "...(a finite union of proper subvector spaces is never a subvector space; details omitted)" is false. A counterexample is any proper subvector space of a vector space. It is sufficient for the lemma to know that a vector space can never be equal to a finite union of proper subspaces if $F$ is infinite.