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The Stacks project

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).

Lemma 110.59.1. The lisse-étale site is not functorial, even for morphisms of schemes.

Proof. See discussion above. \square


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