The Stacks project

7.22 Cocontinuous functors which have a right adjoint

Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ and $v : \mathcal{D} \to \mathcal{C}$ be functors of the underlying categories such that $v$ is right adjoint to $u$. In this case, if $v$ is continuous, then $u$ is cocontinuous (Lemma 7.22.4). If $u$ is cocontinuous, then it is often (but not always, see Example 7.22.5) the case that $v$ is continuous, and if so, then $v$ defines a morphism of sites whose associated morphism of topoi is the same as that defined by $u$.

Lemma 7.22.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$, and $v : \mathcal{D} \to \mathcal{C}$ be functors. Assume that $u$ is cocontinuous and that $v$ is a right adjoint to $u$. Let $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be the morphism of topoi associated to $u$, see Lemma 7.21.1. Then

  1. for a sheaf $\mathcal{F}$ on $\mathcal{C}$ the sheaf $g_*\mathcal{F}$ is equal to the presheaf $v^ p\mathcal{F}$, in other words, $(g_*\mathcal{F})(V) = \mathcal{F}(v(V))$, and

  2. for a sheaf $\mathcal{G}$ on $\mathcal{D}$ we have $g^{-1}\mathcal{G} = (v_ p\mathcal{G})^\# $.

Proof. For $\mathcal{F}$ as in (1) we have

\[ g_*\mathcal{F} = {}_ su\mathcal{F} = {}_ pu\mathcal{F} = v^ p\mathcal{F} = \mathcal{F} \circ v \]

The first equality is Lemma 7.21.1. The second equality is Lemma 7.20.2. The third equality is Lemma 7.19.3. The final equality is the definition of $v^ p$ in Section 7.5. This proves (1). For $\mathcal{G}$ as in (2) we have

\[ g^{-1}\mathcal{G} = (u^ p\mathcal{G})^\# = (v_ p\mathcal{G})^\# \]

The first equality is Lemma 7.21.1. The second equality is Lemma 7.19.3. $\square$

Lemma 7.22.2. Notation and assumptions as in Lemma 7.22.1. If in addition $v$ is continuous then $v$ defines a morphism of sites $f : \mathcal{C} \to \mathcal{D}$ whose associated morphism of topoi is equal to $g$.

Proof. We will use the results of Lemma 7.22.1 without further mention. To prove that $v$ defines a morphism of sites $f$ as in the statement of the lemma, we have to show that $v_ s$ is an exact functor (see Definition 7.14.1). Since $v_ s\mathcal{G} = (v_ p\mathcal{G})^\# = g^{-1}\mathcal{G}$ this follows from the fact that $g$ is a morphism of topoi. Then we see that $f^{-1} = v_ s = g^{-1}$ and we find that $f = g$ as morphisms of topoi. $\square$

Example 7.22.3. This example continues the discussion of Example 7.14.3 from which we borrow the notation $\mathcal{C}, \tau , \tau ', \epsilon $. Observe that the identity functor $v : \mathcal{C}_{\tau '} \to \mathcal{C}_\tau $ is a continuous functor and the identity functor $u : \mathcal{C}_\tau \to \mathcal{C}_{\tau '}$ is a cocontinuous functor. Moreover $u$ is left adjoint to $v$. Hence the results of Lemmas 7.22.1 and 7.22.2 apply and we conclude $v$ defines a morphism of sites, namely

\[ \epsilon : \mathcal{C}_\tau \longrightarrow \mathcal{C}_{\tau '} \]

whose corresponding morphism of topoi is the same as the morphism of topoi associated to the cocontinuous functor $u$.

Lemma 7.22.4. Let $\mathcal{C}$ and $\mathcal{D}$ be sites and let $v : \mathcal{D} \to \mathcal{C}$ be a continuous functor. Assume $v$ has a left adjoint $u : \mathcal{C} \to \mathcal{D}$. Then

  1. $u$ is cocontinuous,

  2. the results of Lemmas 7.22.1 and 7.22.2 hold.

In particular, $v$ defines a morphism of sites $f : \mathcal{D} \to \mathcal{C}$.

Proof. Let $U$ be an object of $\mathcal{C}$ and let $\{ V_ j \to u(U)\} _{j \in J}$ be a covering in $\mathcal{D}$. Then $\{ v(V_ j) \to v(u(U))\} _{i \in I}$ is a covering in $\mathcal{C}$. Via the adjunction map $U \to v(u(U))$ we can base change this to a covering $\{ W_ j \to U\} _{j \in J}$ with $W_ j = v(V_ j) \times _{v(u(U))} U$. Denoting $p_ j : W_ j \to v(V_ j)$ the first projection, we obtain maps

\[ u(W_ j) \xrightarrow {u(p_ j)} u(v(V_ j)) \longrightarrow V_ j \]

where the second arrow is the adjunction map. This determines a morphism $\{ u(W_ j) \to u(U)\} \to \{ V_ j \to u(U)\} $ of families of maps with fixed target, showing that $u$ is indeed cocontinuous. The other statements are immediate from Lemmas 7.22.1 and 7.22.2. $\square$

Example 7.22.5. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ and $v : \mathcal{D} \to \mathcal{C}$ be functors of the underlying categories such that $v$ is right adjoint to $u$. Lemma 7.22.4 shows that if $v$ is continuous, then $u$ is cocontinous. Conversely, if $u$ is cocontinuous, then we can't conclude that $v$ is continuous. We will give an example of this phenomenon using the big étale and smooth sites of a scheme, but presumably there is an elementary example as well. Namely, consider a scheme $S$ and the sites $(\mathit{Sch}/S)_{\acute{e}tale}$ and $(\mathit{Sch}/S)_{smooth}$. We may assume these sites have the same underlying category, see Topologies, Remark 34.11.1. Let $u = v = \text{id}$. Then $u$ as a functor from $(\mathit{Sch}/S)_{\acute{e}tale}$ to $(\mathit{Sch}/S)_{smooth}$ is cocontinuous as every smooth covering of a scheme can be refined by an étale covering, see More on Morphisms, Lemma 37.38.7. Conversely, the functor $v$ from $(\mathit{Sch}/S)_{smooth}$ to $(\mathit{Sch}/S)_{\acute{e}tale}$ is not continuous as a smooth covering is not an étale covering in general.


Comments (1)

Comment #9580 by Matthew Emerton on

In the statement of Lem. 09VR, should the morphism of sites be in the other direction, or am I just getting confused?


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.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00XW. Beware of the difference between the letter 'O' and the digit '0'.