Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A^p$ be a group of sheaves on a topological space $X$, let $F$ be the global sections functor $F(A^p) = A^p(X)$. I have to compute the cohomology of the complex

$0\rightarrow A^1(X) \rightarrow A^2(X) \rightarrow A^3(X) \rightarrow \cdots$.

I have to prove that the cohomology groups of this complex are $0$ for $p \geq 1$. Now after some studying and thinking I've arrived at the following conclusion

CONDITION A = Sheaves {$A^p$} are acyclic + (or if you will $\wedge$) CONDITION B (Just some condition) $\Rightarrow$ $H^p(A^*(X)) = 0$ for $p \geq 1$.

Let's say that so far I've taken care of condition B, now the solution to my problem lies right now solely on the sheaves {$A^p$} being acyclic, and here's where my question lies:

Ok let $T$ be the topology on $X$, I want to prove that the sheaves are acyclic by way of proving that they are flabby, a sheaf is flabby when sections $s \in A^p(U)$ for all $U \subseteq T$ extend to $X$ right? Now let's say that this is not the case and the sheaves {$A^p$} are not flabby for $T$, let $S$ be a topology contained in $T$ for which the sheaves {$A^p$} are flabby, $S \subseteq T$ , so if the topological space was ($X,S$) then my problem would have the solution I want no?

  • What I want to know is if I can just choose the topology in $S$ as my topology, that is, change the topology to a coarser topology (for which the sheaves are still sheaves) to solve the problem? How would the solution change and what would that mean? I mean the global sections are still there right?
share|improve this question
    
"I've been asked" – is this homework? Can you give the exact question? I remember you asked a question like this several weeks ago, and now just as then there are little things that don't make sense. You can't choose your topology arbitrarily. If $S$ and $T$ are incomparable then you will not even be able to pull back / push forward sheaves on $(X, T)$ to $(X, S)$. –  Zhen Lin Jun 3 '12 at 7:01
    
Actually, I see you've already asked this on MO, where Steven Landsburg has pointed out some confusions in your question. –  Zhen Lin Jun 3 '12 at 7:04
    
Hey trust me, definitely not homework, I was not asked, that's just how I worded it, I just corrected the confusion, it was a typo, I typed CONDITION A => CONDITION B => Desired outcome, instead of CONDITION A + CONDITION B => Desired outcome, what Steve said at the end "You're of course right that the acyclicty of the complex of global sections is independent of the topology (assuming you've somehow found another topology for which your question makes sense)" tends more towards answering my question. This is the question NOW, I screwed up w/ the typo and lost the chance at a good answer there –  Mario Carrasco Jun 3 '12 at 13:21
    
No, I mean if I find a topology $S$ contained in $T$, $S \subseteq T$ that makes the sheaves flabby (for which the sheaves are still sheaves), I want to know if I could just instead choose that topology and how that would affect a solution –  Mario Carrasco Jun 3 '12 at 13:32
    
I still don't understand what you're trying to do. Your proposition is blatantly false in many cases of interest: for example, I can take $X$ to be a smooth manifold $M$, $A^p = \Omega^p$ the sheaf of smooth differential $p$-forms. Then $\Omega^p$ is flabby, but the complex $\Omega^0 (M) \to \Omega^1 (M) \to \Omega^2 (M) \to \cdots$ is very, very far from acyclic in general. –  Zhen Lin Jun 3 '12 at 16:41

1 Answer 1

I am going to answer your technical question directly, but I doubt this is what you are really looking for.

Let $X$ and $Y$ be topological spaces, and let $f : X \to Y$ be a continuous map. If $1$ is the one-point space, then there are unique continuous maps $g : Y \to Z$, $h : X \to Z$, and $h = g \circ f$. Obviously, $\textbf{AbSh}(1) \simeq \textbf{Ab}$, so we get direct image functors $h_* : \textbf{AbSh}(X) \to \textbf{Ab}$, $g_* : \textbf{AbSh}(Y) \to \textbf{Ab}$, and $f_* : \textbf{AbSh}(X) \to \textbf{AbSh}(Y)$. Again, it is obvious that $g_*$ and $h_*$ are just the respective global sections functors, and we have $h_* = g_* \circ f_*$. So the cohomology functors $H^p(X, -)$ are just the higher direct images / right derived functors $R^p h_*$, and similarly $H^p(Y, -) \cong R^p g_*$.

Lemma. For an abelian sheaf $\mathscr{F}$ on $X$, $R^q f_* \mathscr{F}$ is the sheaf on $Y$ associated with the presheaf $U \mapsto H^q (f^{-1} U, \mathscr{F})$.

Proof. See [Hartshorne, Algebraic Geometry, Ch. III, Prop. 8.1].

Theorem (Grothendieck). Let $\mathscr{F}$ be an abelian sheaf on $X$. There exists a first quadrant spectral sequence $(E_r^{p,q})$, starting on page $2$ with $$E_2^{p,q} = H^p (Y, R^q f_* \mathscr{F})$$ which converges to $H^{p+q} (X, \mathscr{F})$.

Proof. This is the Grothendieck spectral sequence. See [Weibel, An introduction to homological algebra, Thm 5.8.3]. We need to check that injective sheaves on $X$ get pushed forward to acyclic sheaves on $Y$. But an injective sheaf on $X$ is flasque (in the sense of Godement), and flasque sheaves on $X$ are certainly pushed forward to flasque sheaves on $Y$. Then use the fact that flasque sheaves are acyclic.

share|improve this answer
    
Let me see if I can explain myself a bit better now: Let ($X, T$) be a topological space, let $0\rightarrow A^1 \rightarrow A^2 \rightarrow A^3 \rightarrow \cdots$ be a complex of sheaves and let $0\rightarrow A^1(X) \rightarrow A^2(X) \rightarrow A^3(X) \rightarrow \cdots$ be the complex of global sections. I want to compute the cohomology of the complex of global sections. Let's call PROPOSITION A = "ALL sheaves $A^p$ are acyclic", and let's say, let's just say hypothetically that there's this proposition, which we shall call PROPOSITION B –  Mario Carrasco Jun 6 '12 at 17:13
    
, and I tell you (in a manner of speaking, and not because I was asked or it is homework) that PROPOSITION A and PROPOSITION B together make PROPOSITION C = "$H^p(A^*(X)) = 0$ for $p \geq 1$" true, so we have: PROPOSITION A = TRUE + (or if you will $\wedge$) PROPOSITION B = TRUE $\Rightarrow$ PROPOSITION C = TRUE. If both are true, then PROPOSITION C is true. And I also tell you that PROPOSITION B is true, so all you have to do is prove that PROPOSITION A is true, so now the only thing separating me from PROPOSITION C being true is PROPOSITION A, –  Mario Carrasco Jun 6 '12 at 17:16
    
if PROPOSITION A is true then PROPOSITION A + PROPOSITION B are true and PROPOSITION C is true. –  Mario Carrasco Jun 6 '12 at 17:17
    
So what I want to know is, let's say the sheaves $A^p$ are not flabby for topology $T$, but if I found a topology $S$ contained in $T$, $S \subseteq T$, for which the sheaves $A^p$ are flabby (and still sheaves) and hence acyclic, could I use topology $S$ instead of $T$? What would I lose if I did so? What Steve was saying was that the complex of global sheaves is independent of the topology (for a topology that makes sense), so I was like, why not try to find a topology contained in $T$ that makes the sheaves acyclic? I mean the global sections are still there –  Mario Carrasco Jun 6 '12 at 17:17
    
I already answered the technical part of the question. They are related by a Grothendieck spectral sequence. –  Zhen Lin Jun 6 '12 at 17:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.