relation of annihilators on exact sequence 
Let $0\to M' \to M \rightarrow M''\to 0$ be an exact sequence of modules. I want to show that ${\rm Ann}(M)= {\rm Ann}(M')\cap {\rm Ann}(M^{''})$.

The "$\subset$" case I have shown, but I can't show the "$\supset$" case.
 A: This won't be true in general; it is related to the possibility of the short exact sequence being non-split.  
If $M = M'\oplus M,$ then your equation is true [easy exercise].
But consider the simplest example of a non-split short exact sequence, such
as $0 \to \mathbb Z/p \to \mathbb Z/p^2 \to \mathbb Z/p \to 0.$  In this case your question is false.  
In general (i.e. for modules over more general rings), the question of your equation holds in any particular case can be quite delicate.

Note though that obviously $\mathrm{Ann}(M)\mathrm{Ann}(M'') \subset\mathrm{Ann}(M),$ and so combining this with $\mathrm{Ann}(M) \subset \mathrm{Ann}(M') \cap \mathrm{Ann}(M''),$ we find that $V(\mathrm{Ann}(M)) = V(\mathrm{Ann}(M')) \cup V(\mathrm{Ann}(M'')).$ 
One could also think about this in terms of localizing at prime ideals: $M_{\mathfrak p}$ is non-zero if and only at least one of $M'_{\mathfrak p}$ or $M''_{\mathfrak p}$ is non-zero, since localization is exact.
One could also think geometrically, in terms of supports and stalks.
