The Stacks project

@#7343: OK, I added a proof here. But in the future, can you just type in the arguments in latex in the comment box (or email them to me)?

@#7392. In the discussion between Definition 062E and Lemma 062F there is no $M$, so I think the discussion is fine.

In the discussion between Definition 062E and Lemma 062F, I believe that the equivalence is not entirely correct, because elements acting invertibly on $M$ will automatically be $M$-Kozsul-regular but not $M$-regular. But such elements need not be units of $R$: for instance, let $R=k[x]$ a polynomial ring over a field, $f=x$, $M=R/(x-1)$.

Doesn't lemma 062A imply that $K_{\bullet}(\varphi_{fg}')$ should be homotopy equivalent (rather than isomorphic) to the cone of some map of complexes $K_{\bullet}(\varphi_f')[1] \rightarrow K_{\bullet}(\varphi_g')$?

CRing, in (5), is now at https://math.uchicago.edu/~amathew/cr.html

The fourth sentence of the proof (To prove this...) is unclear. Suggestion: "...consider the collection $\mathcal{B}$ of algebraically independent subsets $C$ such that $A\subset C\subset G$".

In the second paragraph of the proof, we should mention that $B_\alpha\subset B$. And I suggest that we should provide a proof for the fact $|\bigcup_{\alpha\in B^\prime} B_\alpha|=|B^\prime|$, actually I just prove $|\bigcup_{\alpha\in B^\prime} B_\alpha|\leq |B^\prime|$ which will suffice.

Proof: Using axiom of choice, we can construct an injection $\bigcup_{\alpha\in B^\prime} B_\alpha\hookrightarrow \bigsqcup_{\alpha\in B^\prime}B_\alpha$ where $\bigsqcup_{\alpha\in B^\prime}B_\alpha$ denotes the disjoint union, by taking an element into one of the $B_\alpha$'s that contains it.

There are injections $\bigsqcup_{\alpha\in B^\prime}B_\alpha\hookrightarrow \bigsqcup_{\alpha\in B^\prime}\mathbb{N}\cong B^\prime \times \mathbb{N}$.

Thus $|\bigcup_{\alpha\in B^\prime} B_\alpha|\leq |B^\prime\times \mathbb{N}|=\max{|B^\prime|,|\mathbb{N}|}=|B^\prime|$ and the result follows.

@7363 I agree with you, and "Y of finite type" shall be "Y is of finite type".

I agree with Alekos on the typo. There seems to be another typo earlier in the same sentence: 'In particular, the being' does not want the 'the'.

Suggested slogan: Hom functors of $\operatorname{Ch}(\mathcal{A})$ respect the homotopy relation.

In the remark preceding 0A3G, it is said that this result improves Grothendieck's result for Noetherian topological spaces 02UX. This confuses me because Noetherian topological spaces are not in general spectral (sometimes they fail to be sober). Can the remark be made more precise?

Just a typo: in the sentence 'Thus $f^{-1}$ thus characterized by the fact ...', the second 'thus' should be 'is'.

In the second proof, shouldn't it be $\mu_i\otimes 1=\psi\circ\lambda_i$ instead of $\lambda_i=\psi\circ (\mu_i\otimes 1)$?

Yes, it is in the SP, see for instance Lemma 01WM.

@#7377: A universal homeomorphism is integral (EGA IV, 18.12.10). So perhaps this should be (resp. already is) in the Stacks Project.

It may be useful to include the statement that the same holds for any $A \rightarrow B$ that is a universal homeomorphism on spectra (sorry if this is already in the Stacks Project but I missed it!). I think this follows from the characterization of Henselian pairs in terms of lifting idempotents.

Hello!

In the proof of (2), showing the surjection of the canonical map to completion, the sequence of equations must have an "f" in the coefficient of x_1. That is, " ... = x-x_0+f e_1 = x-x_0 -f x_1 + f^2 e_2 = ..."

Thanks

In tag 00S1, "φ:P→P is a morphism of presentations from α to α′" should be "φ:P→P' is a morphism of presentations from α to α′".

I think it would be better to indicate where we use the assumption that $f$ is representable by algebraic spaces. I guess this is used to ensure that $X\times_Y T$ is a quasi-compact algebraic space, hence admits an \' etale cover by an affine scheme $U$; the composite $U\to Y$ then factors through some $Y_{\mu}$, and the same is true for $X\times_Y T$ by the sheaf property.

OK. I see. Even though $a\in \mathcal{O}_{X,x}$ can not be lifted to an element in $\Gamma(X,\mathcal{O}_X)$ in general, it can be lifted to an element in $\Gamma(X,\mathcal{K}'_X)$, which makes $ad-cf$ well-defined.