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.



Showing morphism of sheaves is zero
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.


