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Suppose I have a triangle

$$A \to B \to C \to A[1]$$

in $D(Ab(X))$, the derived category of abelian sheaves on some topological space $X$. For each $x \in X$, there is an exact functor $D(Ab(X)) \to D(Ab)$ that takes a complex of sheaves to the complex of stalks at $x$.

Is my triangle an exact triangle if the image under the stalk functor is exact for all $x \in X$?

I suspect that the triangle need not be exact.

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    $\begingroup$ Take a non-split exact triangle in $D(Ab(X))$ taking values in $\mathbb{Q}$-vector spaces. Change the sign of one of the three morphisms so that it stops being exact. All 'restrictions' to $D(Ab)$ are exact because they actually split therein. $\endgroup$ Mar 19, 2021 at 10:38

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The answer is no, but we can say a bit more : it can become true if you pass to the derived $\infty$-category and replace the words "distinguished triangle" with "cofiber sequence" (modulo the choice of a nullhomotopy)

I'll assume we already know that the composite $A\to C$ vanishes - this cannot be deduced from a stalkwise assumption, as we will see later.

Complete the morphism $A\to B$ to an exact triangle $A\to B\to \mathrm{cofib}\to A[1]$, then there exists a map of sequences of composable morphisms (not triangles a priori)

$\require{AMScd}\begin{CD}A@>>> B@>>> \mathrm{cofib}\\ @VVV @VVV @VVV \\ A @>>> B @>>> C\end{CD}$

(it's not unique, in fact there will be different ones depending on a chosen nullhomotopy of $A\to C$ in the derived $\infty$-category)

I then claim that $\mathrm{cofib}\to C$ is an equivalence: because taking stalks is exact, your assumption implies that it is an equivalence stalkwise.

Further note that $H_n(A_x)\cong H_n(A)_x$ naturally in $A$, so that the assumption implies $H_n(A)_x\cong H_n(B)_x$ for all $n$, and so $H_n(A) \cong H_n(B)$ for all $n$, i.e. the map $\mathrm{cofib}\to C$ is an equivalence.

What we have proved is :

Under your assumptions, there exists a morphism $C\to A[1]$ such that the triangle $A\to B\to C\to A[1]$ is distinguished.

However, as is maybe clear from the proof, this morphism $C\to A[1]$ need not be the one you started with, and in particular it is not clear that your original triangle will be distinguished.

Here's a counterexample to that effect: over $X= Spec(\mathbb Z)$ for instance, find an extension of abelian groups $0\to A\to B\to C\to 0$ which splits when localized at each prime (and thus rationnally), but doesn't split integrally. Then $A\to B\to C\to A[1]$ where we put the $0$ map as the map $C\to A[1]$ is an exact triangle at each stalk, but is not an exact triangle.

The original triangle might not be distinguished.

However, as explained before, there is a map $C\to A[1]$ making it into an exact triangle.

So that's the best we can hope for, and it is true.

(for an explicit counterexample, on can use an extension of $\mathbb Q$ by $\mathbb Z$ given by the following composite morphism $\mathbb{Q\to Q/Z \to Z}/p^\infty \to \mathbb{Q/Z}$ where we use the decomposition of $\mathbb{Q/Z}$ as the sum of its local torsions; where we fix a prime $p$, e.g. $p=2$)

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  • $\begingroup$ How are you getting the map of triangles? TR3 requires both triangles to be exact. $\endgroup$ Mar 18, 2021 at 21:37
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    $\begingroup$ I'm not sure how to phrase it with triangulated categories (not even sure it would work in that setting), but I'm interpreting exact triangles as cofiber sequences in the derived $\infty$-category. That way, any nullhomotopy of the composite $A\to C$ induces a map $\mathrm{cofib} \to C$ making the appropriate diagram commute. Since $A\to C$ is $0$ in the derived category, there exists a null homotopy, and so a map (but again, not a unique one) $\endgroup$ Mar 18, 2021 at 21:42
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    $\begingroup$ @MaximeRamzi How do you check that the map $cofib\to C$ is compaible with the maps $cofib\to A[1]$ and $C\to A[1]$? $\endgroup$ Mar 19, 2021 at 0:43
  • $\begingroup$ @Denis-CharlesCisinski : that's a good point, I don't think I can check that $\endgroup$ Mar 19, 2021 at 7:36
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    $\begingroup$ @MaximeRamzi This means that the answer to the question is negative: if we want to determine if a triangle is distinguished or not, it is not sufficient to check this on stalks. The way you formulate the first half of your answer does not make that clear because you jump to cofiber sequences, which only make sense in the derived $\infty$-category, not in the associated triangulated category. I believe this is clear in your mind but that you should help your readers a little bit more. $\endgroup$ Mar 19, 2021 at 10:17

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