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?

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.

On the other hand, Milne's proof does seem to give functoriality in the case when the morphism is Cartesian, that is, when we're looking at the pairs $(X', X' \setminus (X' \times_X Z)) \to (X,X \setminus Z)$. That's because everything actually appears to work on the big étale site over $X$, so you get it automatically for any scheme $X'$ over $X$.

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.

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

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

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?

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 $H^1_Z(X, \mu_n) \cong H^0(Z,\mathbb{Z}/n)$ 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.

On the other hand, Milne's proof does seem to give functoriality in the case when the morphism is Cartesian, that is, when we're looking at the pairs $(X', X' \setminus (X' \times_X Z)) \to (X,X \setminus Z)$. That's because everything actually appears to work on the big étale site over $X$, so you get it automatically for any scheme $X'$ over $X$.

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?

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.

On the other hand, Milne's proof does seem to give functoriality in the case when the morphism is Cartesian, that is, when we're looking at the pairs $(X', X' \setminus (X' \times_X Z)) \to (X,X \setminus Z)$. That's because everything actually appears to work on the big étale site over $X$, so you get it automatically for any scheme $X'$ over $X$.