Is the cohomology map independent of the inducing morphism?

Fix a category CC. Let F:Cop→SetF:C^{op}\rightarrow \operatorname{Set} be a functor. For any morphism Tu→ST\xrightarrow{u}S such that T×STT\times_ST exists, define
H0(S/T,F)≔ker(F(T)⇉F(T×ST)).
H^0(S/T,F)\mathrel{\unicode{x2254}} \ker(F(T)\rightrightarrows F(T\times_ST)).

Naturally, for another T’\rightarrow ST’\rightarrow S and SS-morphism T’\rightarrow TT’\rightarrow T, this induces a morphism H^0(T/S,F)\rightarrow H^0(T’/S,F)H^0(T/S,F)\rightarrow H^0(T’/S,F).

In Topologies et Faisceaux by Demazure, during the proof of Proposition 1.12, one finds the statement that

By a well-known calculation, the morphism H^0(T/S,F)\rightarrow H^0(T’/S,F)H^0(T/S,F)\rightarrow H^0(T’/S,F) is independent of the morphism T’\rightarrow TT’\rightarrow T.

An attempt (Almost a solution)

We have a commutative diagram:

\begin{matrix}
& & F(S) & & \\
&\swarrow & \downarrow & & \\
H^0(T/S,F) & \rightarrow & F(T) & \rightrightarrows & F(T\times_ST) \\
\downdownarrows & & \downdownarrows & & \downdownarrows \\
H^0(T’/S,F) & \rightarrow & F(T’) & \rightrightarrows & F(T’\times_ST’)
\end{matrix}

\begin{matrix}
& & F(S) & & \\
&\swarrow & \downarrow & & \\
H^0(T/S,F) & \rightarrow & F(T) & \rightrightarrows & F(T\times_ST) \\
\downdownarrows & & \downdownarrows & & \downdownarrows \\
H^0(T’/S,F) & \rightarrow & F(T’) & \rightrightarrows & F(T’\times_ST’)
\end{matrix}

I think the two compositions F(S)\rightarrow F(T)\rightrightarrows F(T’)F(S)\rightarrow F(T)\rightrightarrows F(T’) are naturally equal (we are considering SS-morphisms),
so if we know somehow F(S)\rightarrow H^0(T/S,F)F(S)\rightarrow H^0(T/S,F) is an epimorphism (or surjective as we are all in the category of sets),
then the proof is complete.

Currently I have no idea how to prove this in general; per chance we shall assume that T\rightarrow ST\rightarrow S has a section? Indeed, if it has a section, then I can show that F(S)\rightarrow H^0(T/S,F)F(S)\rightarrow H^0(T/S,F) has a section as well.

Since in the context of Proposition 1.12, the morphism T\rightarrow ST\rightarrow S does have a section, the proof can be said to be complete. But it seems from the text that the result is true in general? (Before this statement one finds the sentence Supposons simplement T\times_STT\times_ST existe,\ldots\ldots)

Thanks in advance for any help or reference.

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1 Answer
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Something is wrong with your link. The text in question is exposأ© IV of SGA 3.

Consider two SS-morphisms f_1,f_2\colon T’ \to Tf_1,f_2\colon T’ \to T. They induce a canonical morphism
\alpha = (f_1,f_2)\colon T’\to T\times_S T\alpha = (f_1,f_2)\colon T’\to T\times_S T
such that
f_1 = p_1 \circ \alpha\quad \text{and} \quad f_2 = p_2 \circ \alpha.f_1 = p_1 \circ \alpha\quad \text{and} \quad f_2 = p_2 \circ \alpha.
I denote by p_1,p_2p_1,p_2 the two canonical arrows T\times_S T\to TT\times_S T\to T.

\mathcal{F}\mathcal{F} induces a morphism
\mathcal{F} (\alpha)\colon \mathcal{F} (T\times_S T) \to \mathcal{F} (T’),\mathcal{F} (\alpha)\colon \mathcal{F} (T\times_S T) \to \mathcal{F} (T’),
which satisfies
\mathcal{F} (f_1) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_1) \quad\text{and}\quad \mathcal{F} (f_2) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_2).\mathcal{F} (f_1) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_1) \quad\text{and}\quad \mathcal{F} (f_2) = \mathcal{F} (\alpha)\circ \mathcal{F} (p_2).

Now H^0 (T/S, F)H^0 (T/S, F) is by definition the equalizer of \mathcal{F} (p_1)\mathcal{F} (p_1) and \mathcal{F} (p_2)\mathcal{F} (p_2), so in particular \mathcal{F} (f_1)\mathcal{F} (f_1) and \mathcal{F} (f_2)\mathcal{F} (f_2) coincide on H^0 (T/S, F)H^0 (T/S, F) by the above identities.

  

 

Thanks for pointing out the error about the link; I have fixed the link now. And thanks for this amazing proof! Such beauty. 🙂
– awllower
yesterday

  

 

@awllower Now that you link to the new edition… note that it actually has exactly the same proof there in a footnote! I remember reading that exposé in the original SGA3 text though… The proof is very natural anyways. And it seems like the new edition has a lot of useful remarks.
– Alejo
yesterday

  

 

I didn’t notice the edition difference! Thanks for the reminder!
– awllower
yesterday

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@awllower Actually, the re-edition of SGA has been a very important and ambitious collaborative project, and I think Société Mathématique de France has published the new versions up to SGA3. The real value of that are the editors’ comments supplied by the experts. Now it seems like SGA4 is in preparation. The TeXed files for SGA4 are also easy to google, but I think it’s still work in progress.
– Alejo
yesterday