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Are the connecting homomorphism induced by Kummer sequence and that of localization sequence commutative?

In other words, is the following statement true?

If it is true, then, how can one prove it?

Please give me any advice.

STATEMENT: Let $X$ be a scheme, $n$ an integer invertible on $X$, $Z$ a closed subscheme of $X$ and $U=X\setminus Z$.

Let $G_{m}$ be an etale sheaf on $X$ defined by sections invertible.

And let $\mu_{X}$ be an etale sheaf on $X$ defined by sections which are n-th roots of 1.

Then, the following two homomorphisms are the same.(Sorry, I don't know how to draw a diagram...)

$H^{i}(U, G_{m}) \stackrel{localization}{\to} H_{Z}^{i+1}(X, G_{m}) \stackrel{Kummer}{\to} H_{Z}^{i+2}(X, \mu_{X})$,

$H^{i}(U, G_{m}) \stackrel{Kummer}{\to} H^{i+1}(U, \mu_{X}) \stackrel{localization} \to H_{Z}^{i+2}(X, \mu_{X})$.

Here, morphisms are connectiong homomorphisms of long exact sequences.

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1 Answer 1

up vote 5 down vote accepted

Let $$0\rightarrow{\cal F}\rightarrow{\cal G}\rightarrow{\cal H}\rightarrow 0$$ be any short exact sequence of sheaves. (In your case, it will be the Kummer sequence.) We can replace ${\cal F,G,H}$ with injective resolutions ${\cal I}^\bullet,{\cal J}^\bullet,{\cal K}^\bullet$ so that we have $${\cal I}^\bullet\rightarrow{\cal J}^\bullet\rightarrow{\cal K}^\bullet$$

Now we have a three-by-three commutative diagram with top row $$ 0\rightarrow \Gamma_Z(X,{\cal I}^\bullet)\rightarrow\Gamma(Z,{\cal I}^\bullet)\rightarrow\Gamma(U,{\cal I}^\bullet|_U) \rightarrow 0$$ and where the middle and bottom rows look the same but with ${\cal J}$ and ${\cal K}$ instead of ${\cal I}$.
Taking cohomology (and using the commutativity of the three-by-three diagram) shows that $$H^i(U,{\cal H})\rightarrow H^{i+1}_Z(X,{\cal H})\rightarrow H^{i+2}_Z(X,{\cal F})$$ is the same as $$H^i(U,{\cal H})\rightarrow H^{i+1}(U,{\cal F})\rightarrow H^{i+2}_Z(X,{\cal F})$$ which is what you asked for.

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Thank you so much! –  Hiro Jun 13 '12 at 15:57

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