Lemma 59.94.1. The map $i^{-1}\beta _{K, \overline{s}, \overline{t}}$ is an isomorphism.

## 59.94 The cospecialization map

Let $f : X \to S$ be a morphism of schemes. Let $\overline{x}$ be a geometric point of $X$ with image $\overline{s} = f(\overline{x})$ in $S$. Let $\overline{t}$ be a geometric point of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Let $K \in D(X_{\acute{e}tale})$. For any morphism $g : Y \to X$ of schemes we write $K|_ Y$ instead of $g_{small}^{-1}K$ and $R\Gamma (Y, K)$ instead of $R\Gamma (Y_{\acute{e}tale}, g_{small}^{-1}K)$. We claim that if

$K$ is bounded below, i.e., $K \in D^+(X_{\acute{e}tale})$,

$f$ is locally acyclic relative to $K$

then there is a *cospecialization map*

which will be closely related to the specialization map considered in Section 59.75 and especially Remark 59.75.8.

To construct the map we consider the morphisms

The unit of the adjunction between $h^{-1}$ and $Rh_*$ gives a map

in $D((X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}}))_{\acute{e}tale})$. Lemma 59.94.1 below shows that the pullback $i^{-1}\beta _{K, \overline{s}, \overline{t}}$ is an isomorphism under the assumptions above. Thus we can define the cospecialization map as the composition

**Proof.**
The construction of the maps $h$, $i$, $\beta _{K, \overline{s}, \overline{t}}$ only depends on the base change of $X$ and $K$ to $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Thus we may and do assume that $S$ is a strictly henselian scheme with closed point $\overline{s}$. Observe that the local acyclicity of $f$ relative to $K$ is preserved by this base change (for example by Lemma 59.93.4 or just directly by comparing strictly henselian rings in this very special case).

Let $\overline{x}$ be a geometric point of $X_{\overline{s}}$. Or equivalently, let $\overline{x}$ be a geometric point whose image by $f$ is $\overline{s}$. Let us compute the stalk of $i^{-1}\beta _{K, \overline{s}, \overline{t}}$ at $\overline{x}$. First, we have

since pullback preserves stalks, see Lemma 59.36.2. Since we are in the situation $S = \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ we see that $h : X_{\overline{t}} \to X$ has the property that $X_{\overline{t}} \times _ X \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{X, \overline{x}}) = F_{\overline{x}, \overline{t}}$. Thus we see that

where the equal sign is Theorem 59.53.1. It follows that the map $(\beta _{K, \overline{s}, \overline{t}})_{\overline{x}}$ is none other than the map $\alpha _{K, \overline{x}, \overline{t}}$ used in Definition 59.93.1. The result follows as we may check whether a map is an isomorphism in stalks by Theorem 59.29.10. $\square$

The cospecialization map when it exists is trying to be the inverse of the specialization map.

Lemma 59.94.2. In the situation above, if in addition $f$ is quasi-compact and quasi-separated, then the diagram

is commutative.

**Proof.**
As in the proof of Lemma 59.94.1 we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Then our maps simplify to $h : X_{\overline{t}} \to X$, $i : X_{\overline{s}} \to X$, and $\beta _{K, \overline{s}, \overline{t}} : K \to Rh_*(K|_{X_{\overline{t}}})$. Using that $(Rf_*K)_{\overline{s}} = R\Gamma (X, K)$ by Theorem 59.53.1 the composition of $sp$ with the base change map $(Rf_*K)_{\overline{t}} \to R\Gamma (X_{\overline{t}}, K)$ is just pullback of cohomology along $h$. This is the same as the map

Now the map $cosp$ first inverts the $=$ sign in this displayed formula, then pulls back along $i$, and finally applies the inverse of $i^{-1}\beta _{K, \overline{s}, \overline{t}}$. Hence we get the desired commutativity. $\square$

Lemma 59.94.3. Let $f : X \to S$ be a morphism of schemes. Let $K \in D(X_{\acute{e}tale})$. Assume

$K$ is bounded below, i.e., $K \in D^+(X_{\acute{e}tale})$,

$f$ is locally acyclic relative to $K$,

$f$ is proper, and

$K$ has torsion cohomology sheaves.

Then for every geometric point $\overline{s}$ of $S$ and every geometric point $\overline{t}$ of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ both the specialization map $sp : (Rf_*K)_{\overline{s}} \to (Rf_*K)_{\overline{t}}$ and the cospecialization map $cosp : R\Gamma (X_{\overline{t}}, K) \to R\Gamma (X_{\overline{s}}, K)$ are isomorphisms.

**Proof.**
By the proper base change theorem (in the form of Lemma 59.91.13) we have $(Rf_*K)_{\overline{s}} = R\Gamma (X_{\overline{s}}, K)$ and similarly for $\overline{t}$. The “correct” proof would be to show that the argument in Lemma 59.94.2 shows that $sp$ and $cosp$ are inverse isomorphisms in this case. Instead we will show directly that $cosp$ is an isomorphism. From the discussion above we see that $cosp$ is an isomorphism if and only if pullback by $i$

is an isomorphism in $D^+(\textit{Ab})$. This is true by the proper base change theorem for the proper morphism $f' : X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}}) \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ by the morphism $\overline{s} \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ and the complex $K' = Rh_*(K|_{X_{\overline{t}}})$. The complex $K'$ is bounded below and has torsion cohomology sheaves by Lemma 59.78.2. Since $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ is strictly henselian with $\overline{s}$ lying over the closed point, we see that the source of the displayed arrow equals $(Rf'_*K')_{\overline{s}}$ and the target equals $R\Gamma (X_{\overline{s}}, K')$ and the displayed map is an isomorphism by the already used Lemma 59.91.13. Thus we see that three out of the four arrows in the diagram of Lemma 59.94.2 are isomorphisms and we conclude. $\square$

Lemma 59.94.4. Let $f : X \to S$ be a morphism of schemes. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. Assume

$f$ is smooth and proper

$\mathcal{F}$ is locally constant, and

$\mathcal{F}_{\overline{x}}$ is a torsion group all of whose elements have order prime to the residue characteristic of $\overline{x}$ for every geometric point $\overline{x}$ of $X$.

Then for every geometric point $\overline{s}$ of $S$ and every geometric point $\overline{t}$ of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ the specialization map $sp : (Rf_*\mathcal{F})_{\overline{s}} \to (Rf_*\mathcal{F})_{\overline{t}}$ is an isomorphism.

**Proof.**
This follows from Lemmas 59.94.3 and 59.93.3 and Proposition 59.93.2.
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)