The Stacks project

Typo: In the proof, last sentence, we didn't define what $W'$ is. It should be $\operatorname{Ext}^1(Z',A)$ instead of $\operatorname{Ext}^1(W',A)$.

A notational issue: if we denote $[Z']:C\to B[1]$ and $[Z]:B\to A[1]$ as explained after Definition 13.27.4, shouldn't we write $[Z][1]\circ [Z']$ instead of $[Z]\circ [Z']$ in the statement and the proof? (Alternatively, $[Z][-1]\circ [Z'][-2]:C[-2]\to A$, by the isomorphism $\operatorname{Hom}_{D(\mathcal{A})}(C,A[2])=\operatorname{Hom}_{D(\mathcal{A})}(C[-2],A)$).

The inclusion of prime ideals is reversed in the second paragraph.

@#8418 I mean in the proof of 13.18.9 (sorry, I should have written #8418 inside that tag).

S. Mac Lane gave a "proof by smart choice of notation" in his book Homology, III, Theorem 2.1 (second proof). Basically the same proof can also be read in B. Mitchell, Theory of Categories, VII, Theorem 1.5.

In case anyone finds it useful, here I wrote a compact proof of the isomorphism $$A' \amalg_A (E \times_B B') \cong (A' \amalg_A E)\times_B B'.$$ It is just Mac Lane's argument from his book Homology plus some more details.

I think there's a typo in the introduction: the Grothendieck group should be subject to the relations $[\mathcal{E}]=[\mathcal{E}'] + [\mathcal{E}'']$.

I think in Lemma 0B1J (2) it should be "... at the generic point of $T$" instead of $Z$; in the proof: "... intersections are proper" instead of transversal.

This should be "chapters" in the plural.

It seems common enough to say "the fiber product of two reduced schemes over a perfect field is reduced". This follows from, say, 020I + 035Z, but maybe it can get its own tag?

It seems that the proof never uses that $Y$ is noetherian.

Minor typo in the introduction: "...not given by an automorphism between..." should probably be ...not given by an isomorphism between..."

Hi. I think it is worth adding the following result in this section or in the smooth ring maps section (I couldn't find it elsewhere).

Let $R$ be a ring and $S$ a smooth $R$-algebra. Assume that $A$ is an $R$-algebra and $(A,I)$ is a henselian pair. Then the homomorphism $Hom_R(S,A)\to Hom_R(S,A/I)$ is surjective.

Proof. This follows from Lemma 00T4, Lemma 09XI and Lemma 07M7.

Of course it can be stated in the style that is written here. I also suggest the slogan "smooth algebras have the lifting property for henselian pairs".

In order to prove (5), if you apply (4) as it is stated, I think you only get that $\mathcal{N}$ and $\mathcal{L}$ are isomorphic, not that the given morphism is an isomorphism. However, from the proof of (4) one sees that the following stronger claim holds: if $\deg(\mathcal{L})\leq 0$ and $s$ is a nonzero section of $\mathcal{L}$, then the morphism $\mathcal{O}_X\to\mathcal{L}$ mapping $1$ to $s$ is an isomorphism. Am I right?

Here is a manifestly canonical construction of the map $f_! \to f_\ast$. Consider the relative diagonal $\Delta : X \to X \times_Y X$ along with the two projections $\pi_i : X \times_Y X \to X$. Since $f$ is separated, $\Delta_! \simeq \Delta_\ast$. Combining this with the tautological transformation $f_! \circ \pi_{2,\ast} \to f_\ast \circ \pi_{1,!}$ gives the desired morphism $$f_! \simeq f_! \circ \pi_{2,\ast} \circ \Delta_\ast \to f_\ast \circ \pi_{1,!} \circ \Delta_\ast \simeq f_\ast \circ \pi_{1,!} \circ \Delta_! \simeq f_\ast.$$

Engelking's first name is misspelled in the bibliography.

Last line of proof: I see y \in f^{-1}(U) \subseteq E, but not y\in V\subseteq E.

Maybe one can use \ref{https://stacks.math.columbia.edu/tag/0B8W} to reduce to the Noetherian case, but I'm not sure if every UFD is a colimit of Noetherian UFDs.

There seems to be at least a typo in the second-to-last sentence. We want to check that the unit map $M_0 \to f_\ast f^\ast M_0$ is an isomorphism. This can be checked after base-changing to $S$ thanks to univeral injectivity. By adjunction, the composition $f^\ast M_0 \to f^\ast f_\ast f^\ast M_0 \to f^\ast M_0$ is the identity, while we have shown that the second map is an isomorphism. Hence $M_0 \to f_\ast f^\ast M_0$ becomes an isomorphism after applying $f^\ast$.

Typo in the line before Lemma 85.11.1: "$a_ n \circ f_\varphi = a_ m$" should be "$a_ m \circ f_\varphi = a_ n$".

I'm a little confused by the notion of simplicial sheaves. Say that $\mathcal{F}$ is a sheaf of $\mathop{\mathit{Sh}}(\mathcal{C}_ {total})$. For each $\varphi:[m]\rightarrow [n]$, there is a comparison morphism $\mathcal{F}(\varphi):\mathcal{F}_ m\rightarrow f_ {\varphi,\ast} \mathcal{F}_ n$. Apply $a_ {m,\ast}$ to $\mathcal{F}(\varphi)$, we get a morphism $a_ {m,\ast}(\mathcal{F}(\varphi)): a_ {m,\ast}\mathcal{F}_ m\rightarrow a_ {n,\ast}\mathcal{F}_ n$ in $\mathop{\mathit{Sh}}(\mathcal{D})$. Hence the construction seems to be a cosimplicial sheaves. Also the morphism $\mathcal{G}_ \bullet \longrightarrow a_ {\bullet , *}\mathcal{F}$ goes the other direction of augmentation in Lemma 14.20.2. So it looks more like a "co-augmentation"?