The Stacks project

Typo: I think in the proof of Lemma 0G8P, part 2, ("Assume (2)..."), M should be replaced by N.

See Lemma 10.131.12 for the statement of the tensor product algebra whose proof also uses a universal property. Are you suggesting to deduce this lemma from that one? What is your justification for the sentence: "It suffices to show that ... when ... is surjective and diagram ... is pullback."?

Hi, I'm a bit confused about this Lemma. It seems to me that $X'$ is (or should be) the normalization of $X_{k'}$, so the diagram displayed above is a diagram that is composed of two smaller diagrams: the top one would be the pushout that produces $X_{k'}$ from its normalization (this is a square of k'-schemes) and the bottom half would be the diagram that one obtains by base changing to $k'$ the line $x\to X\to \textrm{Spec} k$. Am I right? Thanks

This lemma can be deduced from Lemma 17.3.5.

First, it can be shown that: Let $(X,O_X)$, $I$, $\{F_i\}_{i\in I}$ as lemma 17.3.5. Let $E\subset X$ be a quasi-compact subset. Let $U,V\subset X$ are open subset, and assume $U\subset E\subset V$. Then, the map $(\bigoplus F_i)(V)\to (\bigoplus F_i)(U)$ factors through $\bigoplus (F_i(U))$.

Proof. Let $i\colon E\to X$ be a inclusion map. Then $(\bigoplus F_i)(V)\to (\bigoplus F_i)(U)$ factors through $(\bigoplus F_i)|_E(E)$. By lemma 17.3.5, $(\bigoplus F_i)|_E(E)\cong \bigoplus (F_i|_E(E))$. Immediately $\bigoplus (F_i|_E(E))\to (\bigoplus F_i)(U)$ factors through $\bigoplus (F_i(U))$. Hence $(\bigoplus F_i)(V)\to (\bigoplus F_i)(U)$ factors through $\bigoplus (F_i(U))$. q.e.d.

Using it, lemma 17.10.8 can be shown as follows: $F$ is locally isomorphic to the cokernel of $\Psi\colon \bigoplus_{j\in J} O_X\to \bigoplus_{i\in I} O_X$. By assumption, there exists $U\subset E\subset X$ such that U is open and E is quasi-compact. Then $(\bigoplus_{j\in J} O_X)(X)\to (\bigoplus_{i\in I} O_X)(X)$ factors through $\bigoplus_{i\in I} (O_X(U))$. Hence $\Psi|_U$ is given by a matrix. The rest of proof is same as present proof.

You might want to say that $f:X\rightarrow Y$ is a morphism over $S$?

This lemma is essenntially a corollary of 6.29.1, because infinite direct sum is filtered colimit of finite direct sums, and transition maps are injecive.

This lemma can be proved by universal property.

It suffices to show that $\Omega_{S/R} \cong \Omega_{S'/R'}$ when $S\to S'$ is surjective and diagram (10.131.4.1) is pullback.

To show the surjectivity of $\Omega_{S/R} \to \Omega_{S'/R'}$, suppose $d\colon S'\to M$ is a $R'$-derivation and assume $d\circ \phi = 0$. In this case, $d = 0$ because $\phi$ is surjective.

Next, let $d\colon S\to M$ be a $R$-derivation. If $a\in I$, there exists $s\in R$ such that $\alpha(s) = a$. It is because $R = R'\otimes_{S'} S$. Hence $d$ factors through $S'$. By taking $M=\Omega_{S/R}$, we see that $S\to \Omega_{S/R}$ factors through $\Omega_{S'/R'}$. This means that $\Omega_{S/R} \to \Omega_{S'/R'}$ has a retraction. In particular $\Omega_{S/R} \to \Omega_{S'/R'}$ is injective.

We want to have enough weakly contractible objects in our pro-etale site; the function Bound plays a role in the proof of Lemma 61.13.3 (see also Lemmas 61.13.4 and 61.13.5). I did not try to find an optimal function, because the only thing that matters for the later results is the existence of a such a function.

Is is possible to provide a pointer as to why one needs the possibly enlarged function Bound in 098G? For instance, a mention or citation of a step that enlarges the site faster than the default Bound from 046U? I note that the cardinality of the new Bound(κ) is (pointwise) greater than or equal to that of the homset Set(P(P(P(κ)))),κ), so perhaps an indication of why three powersets in particular?

It is not hard, although a bit cumbersome, to verify the claims made; while writing this proof I had numerous sign errors I wrote the scripts to make sure that in the final version no sign errors were present. The link you gave does not prove the claims as far as I can tell.

The verification of $dh-hd=1-\kappa$ is omitted and proved via scripts, which's not elegant as a math proof. I suggest that we follow the proof in this link http://math.stanford.edu/~conrad/papers/cech.pdf.

Thanks.

The sequence (10.138.12.1) is what you get if you apply Lemma 10.138.7 to the formally smooth ring map $R/R \to S/IS$ and the surjection $P/IP \to S/IS$ with kernel $J \otimes_R R/I = J/IJ$ and you use the equalities mentioned in the previous sentences.

According to my understanding, the left hand side quotient in the first short exact sequence (031M) should be

IP + J / (IJ + J^2) instead of J / (IJ + J^2)

It should be K/K^2 where K=ker( P -> S/I)=IP+J. What am I missing?

I was wondering if the last $\mathcal{O}_Y(d')$ should be $\mathcal{O}_X(d')$.

No problem! Yes I think something for categories fibred in groupoids would be good. Maybe what you're saying is true, although it still seems to me that the 2-morphisms are a bit different. I tried for a bit and didn't find a counterexample.

Agreed: argh! Nice simple example! I intend to add this in the future. Thanks!

Interestingly, in your example the category $A$ is not fibred in groupoids over $B$, right? So, maybe what is true: $2$-fibre products in the $(2, 1)$-category of categories fibred in groupoids "agree" with the simlpe one in Example 4.31.3? Then we could add that as a lemma and use it whenever we use the lemmas in this section for categories fibred in groupoids. What do you think?

Obviously, I didn't put enough time into the previous comment and I am not putting a lot of time into this one either. So thank you for helping Anonymous!

Are these fiber products actually equivalent? Let $B$ be the groupoid category with one object and two arrows, and let $A$ be the discrete category with one object. Taking the $2$-fibre product $A \times_B A$ as categories yields the discrete category with two objects. However, if we view all of these as categories over $B$, the $2$-fiber product $A \times_B A$ as categories over $B$ is the discrete category with one object.

The difference is that (in the notation of Lemma 4.32.3), we were allowed to choose any comparison isomorphism $f$ in the first situation, but could only choose the identity arrow in the second situation. I think this reflects the fact that $2$-morphisms in the category of categories over $B$ are a little different.

Also agreed that this is probably some general fact about $2$-fibre products...

I think so. It seems to me the only thing to check is that the $2$-fibre product constructed in Lemma 4.32.3 is canonically equivalent to the one constructed in Lemma 4.31.4. Or can you see other problems? Presumably there is a meta version of each of the lemmas here for any $(2, 1)$ category with $2$-fibre products (and certain products).

Is it clear that this lemma (and other such lemmas in this section) carry over to the $2$-category of categories over categories, as in the next section (Section 4.32)? Or is there an implicit "by the same proof" assertion when the result (for categories over categories) is invoked in Lemma 4.42.6?