Lemma 110.59.1. The lisse-étale site is not functorial, even for morphisms of schemes.
110.59 The lisse-étale site is not functorial
The lisse-étale site $X_{lisse,{\acute{e}tale}}$ of $X$ is the category of schemes smooth over $X$ endowed with (usual) étale coverings, see Cohomology of Stacks, Section 103.14. Let $f : X \to Y$ be a morphism of schemes. There is a functor
\[ u : Y_{lisse,{\acute{e}tale}} \longrightarrow X_{lisse,{\acute{e}tale}},\quad V/Y \longmapsto V \times _ Y X \]
which is continuous. Hence we obtain an adjoint pair of functors
\[ u^ s : \mathop{\mathit{Sh}}\nolimits (X_{lisse,{\acute{e}tale}}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (Y_{lisse,{\acute{e}tale}}), \quad u_ s : \mathop{\mathit{Sh}}\nolimits (Y_{lisse,{\acute{e}tale}}) \longrightarrow \mathop{\mathit{Sh}}\nolimits (X_{lisse,{\acute{e}tale}}), \]
see Sites, Section 7.13. We claim that, in general, $u$ does not define a morphism of sites, see Sites, Definition 7.14.1. In other words, we claim that $u_ s$ is not left exact in general. Note that representable presheaves are sheaves on lisse-étale sites. Hence, by Sites, Lemma 7.13.5 we see that $u_ sh_ V = h_{V \times _ Y X}$. Now consider two morphisms
\[ \xymatrix{ V_1 \ar[rd] \ar@<1ex>[rr]^ a \ar@<-1ex>[rr]_ b & & V_2 \ar[ld] \\ & Y } \]
of schemes $V_1, V_2$ smooth over $Y$. Now if $u_ s$ is left exact, then we would have
\[ u_ s \text{Equalizer}(h_ a, h_ b : h_{V_1} \to h_{V_2}) = \text{Equalizer}(h_{a \times 1}, h_{b \times 1} : h_{V_1 \times _ Y X} \to h_{V_2 \times _ Y X}) \]
We will take the morphisms $a, b : V_1 \to V_2$ such that there exists no morphism from a scheme smooth over $Y$ into $(a = b) \subset V_1$, i.e., such that the left hand side is the empty sheaf, but such that after base change to $X$ the equalizer is nonempty and smooth over $X$. A silly example is to take $X = \mathop{\mathrm{Spec}}(\mathbf{F}_ p)$, $Y = \mathop{\mathrm{Spec}}(\mathbf{Z})$ and $V_1 = V_2 = \mathbf{A}^1_\mathbf {Z}$ with morphisms $a(x) = x$ and $b(x) = x + p$. Note that the equalizer of $a$ and $b$ is the fibre of $\mathbf{A}^1_\mathbf {Z}$ over $(p)$.
Proof. See discussion above. $\square$
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Your email address will not be published.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).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)