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Suppose we have a scheme $X$ and a closed subscheme $Z$, with complement $U$. Then, for any étale sheaf $F$ on $X$, we get a long exact sequence in cohomology

$\cdots H^i(X,F) \to H^i(U,F) \to H^{i+1}_Z(X,F) \to H^{i+1}(X,F) \to \cdots$

as described for example by Theorem 9.4 in Milne's Lecture Notes on Étale Cohomology. (Do we maybe need $Z$ to be reduced?) Milne also says that the sequence is functorial in the pair $(X,X \setminus Z)$. My question is:

What does this mean? In other words, what notion of morphism of pairs makes this statement true?

[Edit: the claimed counterexample below is not one, as pointed out by Ulrich.]

The reason I think clarification is needed is the following example: take $X$ to be the affine line over an algebraically closed field of characteristic zero and $Z$ to be the origin. Then there is a functorial isomorphism $H^2_Z(X, \mu_n) \cong H^0(Z,\mathbb{Z}/n)$ given by the purity theorem, and the exact sequence gives an isomorphism $H^1(U,\mu_n) \to H^0(Z,\mathbb{Z}/n) = \mathbb{Z}/n$. Now let $f \colon X \to X$ be the map given by $x \mapsto x^n$. This seems to induce the zero map on $H^1(U,\mu_n)$ but the identity map on $H^0(Z,\mathbb{Z}/n)$, and so the sequence doesn't seem to be functorial in this instance.

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    $\begingroup$ The indices in your long exact sequence are not right: the third $i$ should be $i+1$. Your purity isomorphism is also wrong. $\endgroup$
    – naf
    Commented May 2, 2014 at 12:07
  • $\begingroup$ @ulrich - thank you, I was being careless. I believe it's correct now. $\endgroup$ Commented May 2, 2014 at 12:13
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    $\begingroup$ One place this is made precise is Bloch--Ogus: Gersten conjecture and the homology of schemes. I don't have time to write more so I'm leaving this as a comment. $\endgroup$ Commented May 2, 2014 at 13:16
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    $\begingroup$ I think if you try to make precise what you mean by "functorial" in the phrase the "functorial isomorphism...given by the purity theorem" you will see what the problem is with your example. $\endgroup$
    – naf
    Commented May 2, 2014 at 14:11
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    $\begingroup$ Cohomology with support is like relative cohomology in topology: $H^i_Z(X,F)$ corresponds to $H^i(X,X−Z;F)$ (for $F$ a constant sheaf). So the funtoriality that one should expect (and which does hold) is for maps of pairs $(X,X−Z)$. $\endgroup$
    – naf
    Commented May 4, 2014 at 5:19

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