The Stacks project

Although you are right that it is equivalent, it is not completely trivial to see the equivalence. The corresponding result for "geometrically reduced" is Lemma 10.43.3 and the corresponding result for "geometrically irreducible" is Lemma 10.46.3 except that unfortunately it is missing the part where we say it is enough to look for the spectrum of $A \otimes_k \overline{k}$ to be irreducible with $\overline{k}$ equal to either the algebraic closure or the separable algebraic closure. I will add this the next time I go through the comments.

After the second sentence, couldn't you just conclude directly from Tag 15.23.5? (In fact, you don't need to take the image, because exactness on the right would not be needed.)

A possibly helpful equivalent definition: If $A$ is a $k$ algebra, then $A$ is geometrically integral iff $A\otimes_k \overline{k}$ is a domain.

The answer to your last question is no. Everything else you say is fine (except I cannot comment on your state of mind). Suggest reading the text by Lenstra (especially section 3) to clarify things, see reference in Section 55.1.

In the statement of the lemma, should $\mathcal{O}_S$ actually be $\mathcal{O}_Y$?

I'm confused. If you leave out basepoints, then objects in the source category will have nontrivial automorphisms. A natural automorphism of the fibre functor will be required to intertwine with these automorphisms. Does this not force the etale fundamental group as defined above to consist only of the central elements in what should be the fundamental group?

No.

Shouldn't the source category of the fibre functor to be, not finite etale coverings, but finite etale coverings with basepoint?

First, of all, we should add a variant of this lemma where $X$ and $Y$ are flat over $S$ in which case this lemma is a lot easier to prove. Secondly, in the formula $R\Gamma (X, \sigma _{\geq n}\mathcal{F}^\bullet ) \otimes _ A^\mathbf {L} R\Gamma (Y, \mathcal{G}^\bullet )$ there is a typo and $n$ should be $N$. Thirdly, the statement that this lives in degrees $\geq N + b$ needs to be justified by proving a bound on the tor dimension of $R\Gamma (Y, \mathcal{G}^\bullet )$ which follows from the results in the next paragraph.

A couple typos:

You should say that $(V,v) \to (Y,y)$ is an elementary étale neighborhood since you assume Y is unibranch, not geometrically unibranch.

Right after that, you should say $u$ is the unique point of $U$ mapping to $v$ not vice versa.

The comments above will soon be obsolete as I am changing the sign rules in this chapter to be more uniform which causes the "problem" above to be fixed.

In the proof of Lemma:10.5.3 , 4) you don't  need the fact that $M_1$ is finite I think and in the proof of 5) you didn't use the fact that $M_2$ is finite ?

Does this Lemma hold for the completion of the strict Henselisation of the local ring?

Let $\widehat{\mathcal{O}^{sh}}_{X,x}$ be the completion of the strict henselization. Then is the fiber $\nu^{-1}(x)$ in bijection with $X^{\nu} \times_X \mathrm{Spec} (\widehat{\mathcal{O}^{sh}}_{X,x})$.

Does this Lemma hold for the completion of the strict Henselisation of the local ring?

Let $\widehat{\mathcal{O}^{sh}}_X,x$ be the completion of the strict henselization. Then is the fiber $\nu^{-1}(x)$ in bijection with $X^{\nu} \times_X \mathrm{Spec} (\widehat{\mathcal{O}^{sh}}_X,x)$.

Sorry I meant to say pushforward, but still should be exact in the presheaf setting.

Adding to the discussion here, I think another point of view is that pullback of sheaves is forgetting it's a sheaf, pulling back, and sheafifying. And the second and third functor do not contribute in the derived setting, hence the statement. I'm not claiming this is any easier than the proof given of course.

There's a typo in the last line, the second term should be $U(R \to R[x_1,\ldots,x_n],M)$

$R[y_1,\ldots,y_n]$ on the $4$-th line from the bottom should be a $R[y_1,\ldots,y_d]$

What is the morphism $g$ in the proof? Is it supposed to refer to $p$?

Also, the end of the first paragraph seems to conclude that $\def\ast{*}$

$$i_1^*p^\ast([W]) = p_1^\ast(c_1(\mathcal{L}_2) \cap [W]) + p_1^\ast(c_1(\mathcal{L}_2) \cap [W]),$$ which is different from $p_1^\ast i^\ast([W])$.

I think the following is a typo: "$\psi_j$" denotes both the maps $V_j\rightarrow U$ in the covering $\mathcal{V}$ as well as the isomorphisms $V_j\rightarrow U_{\beta(j)}$.