I do not see is how to prove that is defined at . The problem, I think, is the following. Let be a quasi-isomorphism. Then I think the arguments in the proof and Lemma 13.5.8 allow us to conclude that given an integer there is a quasi-isomorphism such that the composition is an isomorphism on cohomology in degrees . But in order to show that is defined at with value we would need to show that we can do this in all degrees simultaneously!

Thanks for the interesting question.

]]>And we don't really need the injection in this proof. For the shortest suggestion on the proof, I would add something like:

We proved every element of can be lifted to an element in for some open affine . Providing is functorial. Then the injectivity of on reduces to the injectivity of on , which is already proved.

]]>@#5880. Yes. I'm sorry about the mismatch between the editor and what readers get to see. As you say, this has been a problem for a while. We don't quite know how to fix this.

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