OK. Fixed here.

Excellent catch! I would say this was a bit worse than just a typo. I improved the statement and proof of this lemma a little bit here.

Dear ZL, thank you for the question and apologies for the late reply. Really the category of topoi (and/or of sites) is a 2-category (just like the category of categories is a 2-category). So for some of the diagrams in this section, we shouldn't say that they commute but really that they commute up to a 2-arrow. In the Stacks project we are a bit laks about this and we often use the equal sign "$=$" between functors to indicate that there exists a canonical identification between them, i.e., that there is a canonical invertible 2 arrow from one to the other. Here the word "canonical" indicates that we're too lazy to pin down the isomorphism more precisely.

This works rather well up to a point. If more precision is needed (say beyond 2 catgories into 3 or higher categories), then the best thing is probably to work with $\infty$-categories which are off limits in the Stacks project.

Sure. Change is here.

I don't know if the following is standard knowledge or not (I just found out by myself), but maybe after Definition 22.26.1 the following is an interesting remark to make:

If $\mathcal{A}$ is a differential graded category over $R$, then for all $X\in\operatorname{Ob}\mathcal{A}$ the identity $1_X$ is homogeneous of degree $0$ and $d(1_X)=0$.

Indeed, decompose $1_X=\sum_{n\in\mathbb{Z}}1_{X,n}$ in its homogeneous components. Then $f=f\circ 1_X=\sum_{n\in\mathbb{Z}}f\circ 1_{X,n}$ for $f\in\operatorname{Hom}_\mathcal{A}(X,Y)$. If $f$ is homogeneous, then $f\circ 1_{X,n}$ is homogeneous of degree $\deg f +n$. Thus $f\circ 1_{X,n}=0$ for all $n\neq 0$ and all homogeneous $f\in\operatorname{Hom}_\mathcal{A}(X,Y)$. Therefore $f\circ 1_{X,n}=0$ for all $n\neq 0$ and all $f\in\operatorname{Hom}_\mathcal{A}(X,Y)$. In particular, for $n\neq 0$, we have $1_{X,n}=1_{X,n}\circ 1_X=0$. Hence $1_X=1_{X,0}$ is homogeneous of degree $0$. On the other hand, Leibniz rule gives $d(1_X)=d(1_X\circ 1_X)=d(1_X)\circ 1_X+1_X\circ d(1_X)=2d(1_X)$, whence $d(1_X)=0$.

Yes, we could do this. Then we should also move sections 26.2, 26.3, 26.4 to the chapter on ringed spaces and move any other lemmas etc on locally ringed spaces there as well... On the to-do list.

Going to leave as is.

OK, I added the check that $g$ is qcqs here. Thanks.

Well, to the extent that $R_f$ is not a subring of $R_\mathfrak p$ I agree that the $a_i$ in the first instance are different from the $a_i$ in the second instance. But if $R$ is a domain, say, then $R_f$ may be viewed as a subring of $R_\mathfrak p$ and in that case you can take those $a_i$ to be the same. Anyway, since the sentence is not claiming these $a_i$ are the same, I think I am going to leave this as is until others complain.

OK, I added the description. Thanks to you both! See here.

OK, I finally fixed the closed immersion thing by just proving it affine locally. See here.

Going to leave this as is for now.

Thanks and fixed here.

Going to leave as is for now.

Thanks and fixed here.

Going to leave as is.

8435, #8438, #8439, #8448. You are right that we can define the functor on the slice category (of course) and perhaps we should have done so. You are also right that it is easy to switch between the 2 versions, so I don't see an urgency in making changes. One of the problems with the material in this chapter is the inumerable identifications that are constantly used between different functors / sheaves and it is a bit hard to succintly tell the reader how to think about it. In hindsight, it is best to have as little as possible of this type of material.

OK, the proof was correct as written but it was confusing because of the mix up of the different sites used in the statement and proof. The key is just Lemma 66.16.7 and the fact that in the topology on $X_{spaces, etale}$ the surjective etale morphisms are coverings. See changes here.

Thanks and fixed here.

Thanks and fixed here.