I work in derived category $D^b(X)$ of constructible sheaves on a reasonable space $X$. Let $j\colon U\to X$ be an open inclusion and $i\colon Y\to X$ the closed complement. Let $M,N\in D^b(X)$ and let $f\in Hom_{D^b(X)}(M,N)$. Suppose $i^*f = 0$ and $j^*f= 0$.

Then is it true that $f=0$?

My gut answer is yes, and I thought I would be able to lever the canonical distinguished triangle $j_! j^* \to id \to i_* i^* \to j_!j^*[1]$ into a proof. But I have failed so far.

  • 1
    $\begingroup$ I think a sheaf on $X$ is equivalent to a sheaf $\mathcal{F}$ on $U$, a sheaf $\mathcal{G}$ on $Y$, and a morphism $i^*j_*\mathcal{F} \to \mathcal{G}$. This implies your claim, right? $\endgroup$ – Justin Campbell Sep 30 '12 at 16:43
  • $\begingroup$ These are complexes of sheaves in the derived category, so I am especially skeptical of your claim. Not saying it's wrong, just don't see why, care to elaborate? $\endgroup$ – Reladenine Vakalwe Sep 30 '12 at 17:01

In fact, it is not true. A counterexample is gotten by taking $X = S^1$, $Y$ to be any point on $S^1$, "sheaves" to be, say, sheaves of $\bf{Q}$-vector spaces, and taking $M = \bf{Q}$, $N = \bf{Q} [1]$ and $f$ to be any map classifying a nontrivial first cohomology class on $S^1$. The problem, of course, is that the first cohomologies of $U$ and $Y$ are both trivial.

By the way, I believe that Justin's claim in the comments is true (except that the arrow goes the other way, $G \rightarrow i^*j_*F$), but it needs to be taken in an $\infty$-categorical sense, so that in using it to make maps of sheaves one has extra freedom: choosing a homotopy which makes the requisite diagram commute, rather than simply requiring the diagram to commute in the homotopy category. This applies even when the constituent maps $F\rightarrow F'$ and $G\rightarrow G'$ are zero.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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