Let $\mathcal{C}$ be a small finitely complete category equipped with a Grothendieck (pre)topology $\tau$. For $\ast$ the terminal category (one object, one morphism), denote by $i: \ast \to \mathcal{C}$ the inclusion of the terminal object into $\mathcal{C}$. Equip $\ast$ with the trivial topology in which the only covering is the identity morphism of the unique object onto itself.

By the usual yoga of functoriality of presheaves, $i$ induces a triple of adjoint functors $i_! \dashv i^\ast \dashv i_\ast$ between the categories of presheaves on $\mathcal{C}$ and $\ast$. Trivially $i$ preserves coverings and is left exact —that is, it furnishes a morphism of sites. Then $i^\ast$ takes sheaves to sheaves and so the pair $i_! \dashv i^\ast$ descends to an adjoint pair, $a i_! \dashv i^*$ ($a$ denoting sheafification), of functors between the appropriate categories of sheaves, which is in fact nothing but the geometric morphism of global sections of $\mathrm{Sh}(\mathcal{C}, \tau)$ consisting of the global sections functor $\Gamma: \mathsf{Sh}(\mathcal{C}, \tau) \to \mathsf{Set}$ and its left adjoint, $\mathrm{const}: \mathsf{Set} \to \mathsf{Sh}(\mathcal{C}, \tau)$, taking a set $A$ to the locally constant sheaf on $A$.

It seems to me that $i$ is also a cocontinuous functor (in the terminology of SGA4 or the Stacks Project, section 19; other texts, such as MacLane-Moerdijk [VII.10.Theorem 5] call it a functor having the covering lifting property), making the above geometric morphism of global sections into a local geometric morphism. The latter means that $i_\ast$ takes sheaves to sheaves (Stacks Project, Lemma 19.2), so that there is a triple of adjoint functors $a i_! \dashv i^\ast \dashv i_\ast$ between the categories of sheaves on $(\mathcal{C}, \tau)$ and $\ast$. However, an explicit calculation (see Stacks Project, section 18) of $i_\ast$ yields that $i_\ast(A)$ is the constant presheaf on $A$, which is not in general a sheaf.

I am clearly missing something, but I cannot fathom what it is. Can anybody?


1 Answer 1


The functor $i$ does not have the cover lifting property in general. If it did, then every epimorphism $X \to 1$ in $\mathbf{Sh}(\mathcal{C}, \tau)$ would be an isomorphism, or equivalently, every $\tau$-cover of $1$ in $\mathcal{C}$ must contain a split epimorphism. (Check the definition of "cover lifting property" carefully!)

  • $\begingroup$ @Zhen: I still don't see where I'm going wrong. I'm interested in $(\mathcal{C}$, \tau)$ being the site of Stein spaces with the etale analytic topology (coverings being jointly surjective collections of local isomorphisms). In this case the terminal object is a point with the complex numbers as the sections of its structure sheaf, and it seems to me every cover contains a split epi. $\endgroup$ Jun 18, 2013 at 1:12
  • 1
    $\begingroup$ Then there is no problem, but it seems to me likely that you have miscalculated what $i_*$ does. It should be given by $(i_* A) (C) = A^{\mathcal{C}(1, C)}$, which is not a constant presheaf in general. $\endgroup$
    – Zhen Lin
    Jun 18, 2013 at 6:47
  • $\begingroup$ @Zhen: oh, I see now where I made a mistake in the calculation of $i_\ast$. Thanks for your help! $\endgroup$ Jun 18, 2013 at 12:47

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