The Stacks project

Condition (2) isn't necessary in the presence of (3): in a neighbourhood of $x$ the morphism $f$ is of finite type by (3) combined with Lemma 29.15.8 (and the trivial fact that etale morphisms are of finite type).

"Essentially smooth" is used here and in the next two lemmas. I guess it means formally smooth and essentially of finite type, or maybe essentially of finite presentation, or (equivalently?) it's a localization of a smooth map; but the definition doesn't seem to be in the stacks project.

Minor typo, in the paragraph below equation 10.132.0.1, I think you mean "generated by elements of the form" and not "generated by the element"

Between "Pick $x\in X$" and "$M$ be an acyclic complex", period and "Let" are needed.

Additional ) needed in the most left hand side of the first equation

35.34.1 I cannot see why $\Delta^* \varphi$ is the identity of $V$. Could you please explain why this is true? Thanks.

Ah, I just saw that the answer is given in Grothendieck's TDTE1 (Sem. Bourbaki 190), Section A 1, paragraph d), page 190-06.

In the lemma, presumably the whole discussion is about the naive contangent complex NL_{S/R}. But the text goes back and forth between that and the full cot cmplx L_{S/R}.

Granted, these complexes have the same H_1 (or H^1, as later, in \ref{https://stacks.math.columbia.edu/tag/08R6/cite}), but it might be better to clarify.

At the end of the paragraph starting with "Proof of statement (2)," $H^i$ needs one more $)$. Also, the last $n$ is the proof should be $N$.

Maybe the last two references to 07KX should be to 07KY

In the sentence before the proposition, one ) is redundant in PAb(C)).

Also, V is not necessarily affine in the first place.

"Since $V$ is affine, we have $H^i(V,¥mathcal{F}) = 0$ for $i¥geq 0$" is not true for $i= 0$.

The base case $n=1$ of Lemma 0CKL need not use we have an identity element in $G$, but rather that $\chi$ is taking values in a field.

According to tag 0014, categories in Stacks Project are small unless stated otherwise. That's why I think sets of subobjects are always available for a general abelian category in the sense of Stacks Project.

Typo: "$j_!(\mathcal{E} \xrightarrow {e} \mathcal{F}) = (\mathcal{E} \xrightarrow {s \circ e} \mathcal{G})$" should be "$j_!(\mathcal{E} \xrightarrow {e} \mathcal{G}) = (\mathcal{E} \xrightarrow {s \circ e} \mathcal{F})$"

In Lemma 29.47.9(3), do you mean that $\pi$ is a universal heomomorphism instead of a universal homomorphism? Similarly for Lemma 29.47.10(3).

In the thrid last line the proof says $I^j$ surjective but they are injective. Also, in the last exact sequence the arrows should point backward.

The statement of part (1) of the lemma is somewhat confusing --- it is better to first put the assumption and then write the conclusion, e.g., "there exist integers $a \le b$ such that for any $L$ with $H^i(L) = 0$ for $i \in [a,b]$ we have $\mathrm{Hom}(K,L) = 0$".

Dear Aise Johan, Sorry. :)

But i must correct myself: i think that the axiom as stated here does imply that $\mathcal F(\emptyset)=0$. The argument though has to be modified: we see that in "there exists a unique section $s\in\mathcal F(\emptyset)$ such that $s_i=s|_U_i$ for all $i\in I$" the restriction condition is always true as $i\in I$ is always false, so the first part says that there exists a section $s\in\mathcal F(\emptyset)$ and that any section $t\in\mathcal F(\emptyset)$ (which necessarily satisfies the always-true restriction condition) equals $s$, thus $\mathcal F(\emptyset)$ is indeed $0$.

Also i should have said that Hartshorne actually imposes all presheaves to have $\mathcal F(\emptyset)=0$, so sheaves a fortiori satisfy this, though his axiom for sheaves would of course enforce this, as above.

And to finish with the equalizer argument: category theorists and logicians surely know when it is possible to translate exactly (even with empty set subtleties) between formulas and diagrams. Here i think it is not trivial to watertight-prove that the set formulation of the axiom does translate into the expected diagrams (in all types of value categories), including all subtleties of vacuous conditions.

I am really sorry to have annoyed you -plus making mistakes in my comment. I can only imagine how tedious this project is -hopefully there are some funnier sections than this one. So i thank you for all your work here and elsewhere. Best wishes, Paul

The point of 6.7.2 was to prevent having this discussion!