# Showing morphism of sheaves is zero

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.

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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? –  Justin Campbell Sep 30 '12 at 16:43
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? –  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.