# Does n-multiplication maps of cohomology groups vanish if it vanishes at the 0th cohomology?

In general, we know that a morphism $f=(f ^ {q})$ between universal (cohomological) $\delta$ functors $S=(S ^ {q}),T=(T ^ {q})\$vanishes if and only if $f ^ {0} \ \colon \ S^{0} \to T^{0}$ vanishes.

However, for a fixed object $F$, we cannot say that $f^{q} (F) = 0$ for every $q$ even if $f^{0} (F) = 0$.

Now, in order to make this statement true, what kind of condition we need for the morphism $f \$?

In particular, I am interested in the following situation:

Let $n$ be an integer, $X$ be a scheme, $F\ \colon \ X _ {et} \to Ab \$ be a $n$-torsion etale sheaf on $X$, and $S=T=H^{\ast}(X,-)$, then can we say that the morphism $H^{q}(X,F)\to H^{q}(X,F)$ induced by the $n$-multiplication map vanishes? Thanks!

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Shouldn't this (meaning your last statement) just follow from functoriality? If by "$\mathcal{F}$ is $n$-torsion" you mean that the multiplication by $n$ map on $\mathcal{F}$ vanishes, then the fact that $H^q(X,\cdot)$ is a functor implies that your map vanishes as well. –  Ramsey Aug 7 '11 at 14:55
Oh, you are right! I was forgetting that any additive functor sends zero morphisms to zero. Thank you so much! –  Hiro Aug 7 '11 at 15:07
Yes, I guess I should have said "additive functor" everywhere above... –  Ramsey Aug 8 '11 at 16:17