
## 7.21 Cocontinuous functors and morphisms of topoi

It is clear from the above that a cocontinuous functor $u$ gives a morphism of topoi in the same direction as $u$. Thus this is in the opposite direction from the morphism of topoi associated (under certain conditions) to a continuous $u$ as in Definition 7.14.1, Proposition 7.14.7, and Lemma 7.15.2.

Lemma 7.21.1. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be cocontinuous. The functors $g_* = {}_ su$ and $g^{-1} = (u^ p\ )^\#$ define a morphism of topoi $g$ from $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ to $\mathop{\mathit{Sh}}\nolimits (\mathcal{D})$.

Proof. This is exactly the content of Lemma 7.20.3. $\square$

Lemma 7.21.2. Let $u : \mathcal{C} \to \mathcal{D}$, and $v : \mathcal{D} \to \mathcal{E}$ be cocontinuous functors. Then $v \circ u$ is cocontinuous and we have $h = g \circ f$ where $f : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$, resp. $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{E})$, resp. $h : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{E})$ is the morphism of topoi associated to $u$, resp. $v$, resp. $v \circ u$.

Proof. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $\{ E_ i \to v(u(U))\}$ be a covering of $U$ in $\mathcal{E}$. By assumption there exists a covering $\{ D_ j \to u(U)\}$ in $\mathcal{D}$ such that $\{ v(D_ j) \to v(u(U))\}$ refines $\{ E_ i \to v(u(U))\}$. Also by assumption there exists a covering $\{ C_ l \to U\}$ in $\mathcal{C}$ such that $\{ u(C_ l) \to u(U)\}$ refines $\{ D_ j \to u(U)\}$. Then it is true that $\{ v(u(C_ l)) \to v(u(U))\}$ refines the covering $\{ E_ i \to v(u(U))\}$. This proves that $v \circ u$ is cocontinuous. To prove the last assertion it suffices to show that ${}_ sv \circ {}_ su = {}_ s(v \circ u)$. It suffices to prove that ${}_ pv \circ {}_ pu = {}_ p(v \circ u)$, see Lemma 7.20.2. Since ${}_ pu$, resp. ${}_ pv$, resp. ${}_ p(v \circ u)$ is right adjoint to $u^ p$, resp. $v^ p$, resp. $(v \circ u)^ p$ it suffices to prove that $u^ p \circ v^ p = (v \circ u)^ p$. And this is direct from the definitions. $\square$

Example 7.21.3. Let $X$ be a topological space. Let $j : U \to X$ be the inclusion of an open subspace. Recall that we have sites $X_{Zar}$ and $U_{Zar}$, see Example 7.6.4. Recall that we have the functor $u : X_{Zar} \to U_{Zar}$ associated to $j$ which is continuous and gives rise to a morphism of sites $U_{Zar} \to X_{Zar}$, see Example 7.14.2. This also gives a morphism of topoi $(j_*, j^{-1})$. Next, consider the functor $v : U_{Zar} \to X_{Zar}$, $V \mapsto v(V) = V$ (just the same open but now thought of as an object of $X_{Zar}$). This functor is cocontinuous. Namely, if $v(V) = \bigcup _{j \in J} W_ j$ is an open covering in $X$, then each $W_ j$ must be a subset of $U$ and hence is of the form $v(V_ j)$, and trivially $V = \bigcup _{j \in J} V_ j$ is an open covering in $U$. We conclude by Lemma 7.21.1 above that there is a morphism of topoi associated to $v$

$\mathop{\mathit{Sh}}\nolimits (U) \longrightarrow \mathop{\mathit{Sh}}\nolimits (X)$

given by ${}_ sv$ and $(v^ p\ )^\#$. We claim that actually $(v^ p\ )^\# = j^{-1}$ and that ${}_ sv = j_*$, in other words, that this is the same morphism of topoi as the one given above. Perhaps the easiest way to see this is to realize that for any sheaf $\mathcal{G}$ on $X$ we have $v^ p\mathcal{G}(V) = \mathcal{G}(V)$ which according to Sheaves, Lemma 6.31.1 is a description of $j^{-1}\mathcal{G}$ (and hence sheafification is superfluous in this case). The equality of ${}_ sv$ and $j_*$ follows by uniqueness of adjoint functors (but may also be computed directly).

Example 7.21.4. This example is a slight generalization of Example 7.21.3. Let $f : X \to Y$ be a continuous map of topological spaces. Assume that $f$ is open. Recall that we have sites $X_{Zar}$ and $Y_{Zar}$, see Example 7.6.4. Recall that we have the functor $u : Y_{Zar} \to X_{Zar}$ associated to $f$ which is continuous and gives rise to a morphism of sites $X_{Zar} \to Y_{Zar}$, see Example 7.14.2. This also gives a morphism of topoi $(f_*, f^{-1})$. Next, consider the functor $v : X_{Zar} \to Y_{Zar}$, $U \mapsto v(U) = f(U)$. This functor is cocontinuous. Namely, if $f(U) = \bigcup _{j \in J} V_ j$ is an open covering in $Y$, then setting $U_ j = f^{-1}(V_ j) \cap U$ we get an open covering $U = \bigcup U_ j$ such that $f(U) = \bigcup f(U_ j)$ is a refinement of $f(U) = \bigcup V_ j$. We conclude by Lemma 7.21.1 above that there is a morphism of topoi associated to $v$

$\mathop{\mathit{Sh}}\nolimits (X) \longrightarrow \mathop{\mathit{Sh}}\nolimits (Y)$

given by ${}_ sv$ and $(v^ p\ )^\#$. We claim that actually $(v^ p\ )^\# = f^{-1}$ and that ${}_ sv = f_*$, in other words, that this is the same morphism of topoi as the one given above. For any sheaf $\mathcal{G}$ on $Y$ we have $v^ p\mathcal{G}(U) = \mathcal{G}(f(U))$. On the other hand, we may compute $u_ p\mathcal{G}(U) = \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{G}(V) = \mathcal{G}(f(U))$ because clearly $(f(U), U \subset f^{-1}(f(U)))$ is an initial object of the category $\mathcal{I}_ U^ u$ of Section 7.5. Hence $u_ p = v^ p$ and we conclude $f^{-1} = u_ s = (v^ p\ )^\#$. The equality of ${}_ sv$ and $f_*$ follows by uniqueness of adjoint functors (but may also be computed directly).

In the first Example 7.21.3 the functor $v$ is also continuous. But in the second Example 7.21.4 it is generally not continuous because condition (2) of Definition 7.13.1 may fail. Hence the following lemma applies to the first example, but not to the second.

Lemma 7.21.5. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Assume that

1. $u$ is cocontinuous, and

2. $u$ is continuous.

Let $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ be the associated morphism of topoi. Then

1. sheafification in the formula $g^{-1} = (u^ p\ )^\#$ is unnecessary, in other words $g^{-1}(\mathcal{G})(U) = \mathcal{G}(u(U))$,

2. $g^{-1}$ has a left adjoint $g_{!} = (u_ p\ )^\#$, and

3. $g^{-1}$ commutes with arbitrary limits and colimits.

Proof. By Lemma 7.13.2 for any sheaf $\mathcal{G}$ on $\mathcal{D}$ the presheaf $u^ p\mathcal{G}$ is a sheaf on $\mathcal{C}$. And then we see the adjointness by the following string of equalities

\begin{eqnarray*} \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, g^{-1}\mathcal{G}) & = & \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{F}, u^ p\mathcal{G}) \\ & = & \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(u_ p\mathcal{F}, \mathcal{G}) \\ & = & \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(g_{!}\mathcal{F}, \mathcal{G}) \end{eqnarray*}

The statement on limits and colimits follows from the discussion in Categories, Section 4.24. $\square$

In the situation of Lemma 7.21.5 above we see that we have a sequence of adjoint functors

$g_{!}, \ g^{-1}, \ g_*.$

The functor $g_!$ is not exact in general, because it does not transform a final object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ into a final object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ in general. See Sheaves, Remark 6.31.13. On the other hand, in the topological setting of Example 7.21.3 the functor $j_!$ is exact on abelian sheaves, see Modules, Lemma 17.3.4. The following lemma gives the generalization to the case of sites.

Lemma 7.21.6. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Assume that

1. $u$ is cocontinuous,

2. $u$ is continuous, and

3. fibre products and equalizers exist in $\mathcal{C}$ and $u$ commutes with them.

In this case the functor $g_!$ above commutes with fibre products and equalizers (and more generally with finite connected limits).

Proof. Assume (a), (b), and (c). We have $g_! = (u_ p\ )^\#$. Recall (Lemma 7.10.1) that limits of sheaves are equal to the corresponding limits as presheaves. And sheafification commutes with finite limits (Lemma 7.10.14). Thus it suffices to show that $u_ p$ commutes with fibre products and equalizers. To do this it suffices that colimits over the categories $(\mathcal{I}_ V^ u)^{opp}$ of Section 7.5 commute with fibre products and equalizers. This follows from Lemma 7.5.1 and Categories, Lemma 4.19.9. $\square$

The following lemma deals with a case that is even more like the morphism associated to an open immersion of topological spaces.

Lemma 7.21.7. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Assume that

1. $u$ is cocontinuous,

2. $u$ is continuous, and

3. $u$ is fully faithful.

For $g_!, g^{-1}, g_*$ as above the canonical maps $\mathcal{F} \to g^{-1}g_!\mathcal{F}$ and $g^{-1}g_*\mathcal{F} \to \mathcal{F}$ are isomorphisms for all sheaves $\mathcal{F}$ on $\mathcal{C}$.

Proof. Let $X$ be an object of $\mathcal{C}$. In Lemmas 7.20.2 and 7.21.5 we have seen that sheafification is not necessary for the functors $g^{-1} = (u^ p\ )^\#$ and $g_{*} = ({}_ pu\ )^\#$. We may compute $(g^{-1}g_{*}\mathcal{F})(X) = g_{*}\mathcal{F}(u(X)) = \mathop{\mathrm{lim}}\nolimits \mathcal{F}(Y)$. Here the limit is over the category of pairs $(Y, u(Y) \to u(X))$ where the morphisms $u(Y) \to u(X)$ are not required to be of the form $u(\alpha )$ with $\alpha$ a morphism of $\mathcal{C}$. By assumption (c) we see that they automatically come from morphisms of $\mathcal{C}$ and we deduce that the limit is the value on $(X, u(\text{id}_ X))$, i.e., $\mathcal{F}(X)$. This proves that $g^{-1}g_{*}\mathcal{F} = \mathcal{F}$.

On the other hand, $(g^{-1}g_{!}\mathcal{F})(X) = g_{!}\mathcal{F}(u(X)) = (u_ p\mathcal{F})^\# (u(X))$, and $u_ p\mathcal{F}(u(X)) = \mathop{\mathrm{colim}}\nolimits \mathcal{F}(Y)$. Here the colimit is over the category of pairs $(Y, u(X) \to u(Y))$ where the morphisms $u(X) \to u(Y)$ are not required to be of the form $u(\alpha )$ with $\alpha$ a morphism of $\mathcal{C}$. By assumption (c) we see that they automatically come from morphisms of $\mathcal{C}$ and we deduce that the colimit is the value on $(X, u(\text{id}_ X))$, i.e., $\mathcal{F}(X)$. Thus for every $X \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we have $u_ p\mathcal{F}(u(X)) = \mathcal{F}(X)$. Since $u$ is cocontinuous and continuous any covering of $u(X)$ in $\mathcal{D}$ can be refined by a covering (!) $\{ u(X_ i) \to u(X)\}$ of $\mathcal{D}$ where $\{ X_ i \to X\}$ is a covering in $\mathcal{C}$. This implies that $(u_ p\mathcal{F})^+(u(X)) = \mathcal{F}(X)$ also, since in the colimit defining the value of $(u_ p\mathcal{F})^+$ on $u(X)$ we may restrict to the cofinal system of coverings $\{ u(X_ i) \to u(X)\}$ as above. Hence we see that $(u_ p\mathcal{F})^+(u(X)) = \mathcal{F}(X)$ for all objects $X$ of $\mathcal{C}$ as well. Repeating this argument one more time gives the equality $(u_ p\mathcal{F})^\# (u(X)) = \mathcal{F}(X)$ for all objects $X$ of $\mathcal{C}$. This produces the desired equality $g^{-1}g_!\mathcal{F} = \mathcal{F}$. $\square$

Finally, here is a case that does not have any corresponding topological example. We will use this lemma to see what happens when we enlarge a “partial universe” of schemes keeping the same topology. In the situation of the lemma, the morphism of topoi $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ identifies $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ as a subtopos of $\mathop{\mathit{Sh}}\nolimits (\mathcal{D})$ (Section 7.43) and moreover, the given embedding has a retraction.

Lemma 7.21.8. Let $\mathcal{C}$ and $\mathcal{D}$ be sites. Let $u : \mathcal{C} \to \mathcal{D}$ be a functor. Assume that

1. $u$ is cocontinuous,

2. $u$ is continuous,

3. $u$ is fully faithful,

4. fibre products exist in $\mathcal{C}$ and $u$ commutes with them, and

5. there exist final objects $e_\mathcal {C} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $e_\mathcal {D} \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{D})$ such that $u(e_\mathcal {C}) = e_\mathcal {D}$.

Let $g_!, g^{-1}, g_*$ be as above. Then, $u$ defines a morphism of sites $f : \mathcal{D} \to \mathcal{C}$ with $f_* = g^{-1}$, $f^{-1} = g_!$. The composition

$\xymatrix{ \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) \ar[r]^ g & \mathop{\mathit{Sh}}\nolimits (\mathcal{D}) \ar[r]^ f & \mathop{\mathit{Sh}}\nolimits (\mathcal{C}) }$

is isomorphic to the identity morphism of the topos $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$. Moreover, the functor $f^{-1}$ is fully faithful.

Proof. By assumption the functor $u$ satisfies the hypotheses of Proposition 7.14.7. Hence $u$ defines a morphism of sites and hence a morphism of topoi $f$ as in Lemma 7.15.2. The formulas $f_* = g^{-1}$ and $f^{-1} = g_!$ are clear from the lemma cited and Lemma 7.21.5. We have $f_* \circ g_* = g^{-1} \circ g_* \cong \text{id}$, and $g^{-1} \circ f^{-1} = g^{-1} \circ g_! \cong \text{id}$ by Lemma 7.21.7.

We still have to show that $f^{-1}$ is fully faithful. Let $\mathcal{F}, \mathcal{G} \in \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}))$. We have to show that the map

$\mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}) \longrightarrow \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G})$

is bijective. But the right hand side is equal to

\begin{align*} \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{D})}(f^{-1}\mathcal{F}, f^{-1}\mathcal{G}) & = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, f_*f^{-1}\mathcal{G}) \\ & = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, g^{-1}f^{-1}\mathcal{G}) \\ & = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \mathcal{G}) \end{align*}

(the first equality by adjunction) which proves what we want. $\square$

Example 7.21.9. Let $X$ be a topological space. Let $i : Z \to X$ be the inclusion of a subset (with induced topology). Consider the functor $u : X_{Zar} \to Z_{Zar}$, $U \mapsto u(U) = Z \cap U$. At first glance it may appear that this functor is cocontinuous as well. After all, since $Z$ has the induced topology, shouldn't any covering of $U\cap Z$ it come from a covering of $U$ in $X$? Not so! Namely, what if $U \cap Z = \emptyset$? In that case, the empty covering is a covering of $U \cap Z$, and the empty covering can only be refined by the empty covering. Thus we conclude that $u$ cocontinuous $\Rightarrow$ every nonempty open $U$ of $X$ has nonempty intersection with $Z$. But this is not sufficient. For example, if $X = \mathbf{R}$ the real number line with the usual topology, and $Z = \mathbf{R} \setminus \{ 0\}$, then there is an open covering of $Z$, namely $Z = \{ x < 0\} \cup \bigcup _ n \{ 1/n < x\}$ which cannot be refined by the restriction of any open covering of $X$.

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