6.29 Limits and colimits of sheaves
Let $X$ be a topological space. Let $\mathcal{I} \to \mathop{\mathit{Sh}}\nolimits (X)$, $i \mapsto \mathcal{F}_ i$ be a diagram.
Both $\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i$ and $\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$ exist.
The inclusion functor $i : \mathop{\mathit{Sh}}\nolimits (X) \to \textit{PSh}(X)$ commutes with limits. In other words, we may compute the limit in the category of sheaves as the limit in the category of presheaves. In particular, for any open $U \subset X$ we have
\[ (\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i)(U) = \mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i(U). \]
The inclusion functor $i : \mathop{\mathit{Sh}}\nolimits (X) \to \textit{PSh}(X)$ does not commute with colimits in general (not even with finite colimits – think surjections). The colimit is computed as the sheafification of the colimit in the category of presheaves:
\[ \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i = \Big(U \mapsto \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(U)\Big)^\# . \]
Let $x \in X$ be a point. In general the stalk of $\mathop{\mathrm{lim}}\nolimits _ i \mathcal{F}_ i$ at $x$ is not equal to the limit of the stalks. But if the index category is finite then it is the case. In other words, the stalk functor is left exact.
Let $x \in X$. We always have
\[ (\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i)_ x = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_{i, x}. \]
The sheafification functor ${}^\# : \textit{PSh}(X) \to \mathop{\mathit{Sh}}\nolimits (X)$ commutes with all colimits, and with finite limits. But it does not commute with all limits.
The proofs are all easy. Here is an example of what is true for directed colimits of sheaves.
Lemma 6.29.1. Let $X$ be a topological space. Let $I$ be a directed set. Let $(\mathcal{F}_ i, \varphi _{ii'})$ be a system of sheaves of sets over $I$, see Categories, Section 4.21. Let $U \subset X$ be an open subset. Consider the canonical map
\[ \Psi : \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(U) \longrightarrow \left(\mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i\right)(U) \]
If all the transition maps are injective then $\Psi $ is injective for any open $U$.
If $U$ is quasi-compact, then $\Psi $ is injective.
If $U$ is quasi-compact and all the transition maps are injective then $\Psi $ is an isomorphism.
If $U$ has a cofinal system of open coverings $\mathcal{U} : U = \bigcup _{j\in J} U_ j$ with $J$ finite and $U_ j \cap U_{j'}$ quasi-compact for all $j, j' \in J$, then $\Psi $ is bijective.
Proof.
Assume all the transition maps are injective. In this case the presheaf $\mathcal{F}' : V \mapsto \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i(V)$ is separated (see Definition 6.11.2). By the discussion above we have $(\mathcal{F}')^\# = \mathop{\mathrm{colim}}\nolimits _ i \mathcal{F}_ i$. By Lemma 6.17.5 we see that $\mathcal{F}' \to (\mathcal{F}')^\# $ is injective. This proves (1).
Assume $U$ is quasi-compact. Suppose that $s \in \mathcal{F}_ i(U)$ and $s' \in \mathcal{F}_{i'}(U)$ give rise to elements on the left hand side which have the same image under $\Psi $. Since $U$ is quasi-compact this means there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$ and for each $j$ an index $i_ j \in I$, $i_ j \geq i$, $i_ j \geq i'$ such that $\varphi _{ii_ j}(s)$ and $\varphi _{i'i_ j}(s')$ have the same restriction to $U_ j$. Let $i''\in I$ be $\geq $ than all of the $i_ j$. We conclude that $\varphi _{ii''}(s)$ and $\varphi _{i'i''}(s')$ agree on the opens $U_ j$ for all $j$ and hence that $\varphi _{ii''}(s) = \varphi _{i'i''}(s')$. This proves (2).
Assume $U$ is quasi-compact and all transition maps injective. Let $s$ be an element of the target of $\Psi $. Since $U$ is quasi-compact there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$, for each $j$ an index $i_ j \in I$ and $s_ j \in \mathcal{F}_{i_ j}(U_ j)$ such that $s|_{U_ j}$ comes from $s_ j$ for all $j$. Pick $i \in I$ which is $\geq $ than all of the $i_ j$. By (1) the sections $\varphi _{i_ ji}(s_ j)$ agree over the overlaps $U_ j \cap U_{j'}$. Hence they glue to a section $s' \in \mathcal{F}_ i(U)$ which maps to $s$ under $\Psi $. This proves (3).
Assume the hypothesis of (4). In particular we see that $U$ is quasi-compact and hence by (2) we have injectivity of $\Psi $. Let $s$ be an element of the target of $\Psi $. By assumption there exists a finite open covering $U = \bigcup _{j = 1, \ldots , m} U_ j$, with $U_ j \cap U_{j'}$ quasi-compact for all $j, j' \in J$ and for each $j$ an index $i_ j \in I$ and $s_ j \in \mathcal{F}_{i_ j}(U_ j)$ such that $s|_{U_ j}$ is the image of $s_ j$ for all $j$. Since $U_ j \cap U_{j'}$ is quasi-compact we can apply (2) and we see that there exists an $i_{jj'} \in I$, $i_{jj'} \geq i_ j$, $i_{jj'} \geq i_{j'}$ such that $\varphi _{i_ ji_{jj'}}(s_ j)$ and $\varphi _{i_{j'}i_{jj'}}(s_{j'})$ agree over $U_ j \cap U_{j'}$. Choose an index $i \in I$ which is bigger or equal than all the $i_{jj'}$. Then we see that the sections $\varphi _{i_ ji}(s_ j)$ of $\mathcal{F}_ i$ glue to a section of $\mathcal{F}_ i$ over $U$. This section is mapped to the element $s$ as desired.
$\square$
Example 6.29.2. Let $X = \{ s_1, s_2, \xi _1, \xi _2, \xi _3, \ldots \} $ as a set. Declare a subset $U \subset X$ to be open if $s_1 \in U$ or $s_2 \in U$ implies $U$ contains all of the $\xi _ i$. Let $U_ n = \{ \xi _ n, \xi _{n + 1}, \ldots \} $, and let $j_ n : U_ n \to X$ be the inclusion map. Set $\mathcal{F}_ n = j_{n, *}\underline{\mathbf{Z}}$. There are transition maps $\mathcal{F}_ n \to \mathcal{F}_{n + 1}$. Let $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}_ n$. Note that $\mathcal{F}_{n, \xi _ m} = 0$ if $m < n$ because $\{ \xi _ m\} $ is an open subset of $X$ which misses $U_ n$. Hence we see that $\mathcal{F}_{\xi _ n} = 0$ for all $n$. On the other hand the stalk $\mathcal{F}_{s_ i}$, $i = 1, 2$ is the colimit
\[ M = \mathop{\mathrm{colim}}\nolimits _ n \prod \nolimits _{m \geq n} \mathbf{Z} \]
which is not zero. We conclude that the sheaf $\mathcal{F}$ is the direct sum of the skyscraper sheaves with value $M$ at the closed points $s_1$ and $s_2$. Hence $\Gamma (X, \mathcal{F}) = M \oplus M$. On the other hand, the reader can verify that $\mathop{\mathrm{colim}}\nolimits _ n \Gamma (X, \mathcal{F}_ n) = M$. Hence some condition is necessary in part (4) of Lemma 6.29.1 above.
There is a version of the previous lemma dealing with sheaves on a diagram of spectral spaces. To state it we introduce some notation. Let $\mathcal{I}$ be a cofiltered index category. Let $i \mapsto X_ i$ be a diagram of spectral spaces over $\mathcal{I}$ such that for $a : j \to i$ in $\mathcal{I}$ the corresponding map $f_ a : X_ j \to X_ i$ is spectral. Set $X = \mathop{\mathrm{lim}}\nolimits X_ i$ and denote $p_ i : X \to X_ i$ the projection.
Lemma 6.29.3. In the situation described above, let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and let $\mathcal{G}$ be a sheaf on $X_ i$. For $U_ i \subset X_ i$ quasi-compact open we have
\[ p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_ a^{-1}\mathcal{G}(f_ a^{-1}(U_ i)) \]
Proof.
Let us prove the canonical map $\mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_ a^{-1}\mathcal{G}(f_ a^{-1}(U_ i)) \to p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i))$ is injective. Let $s, s'$ be sections of $f_ a^{-1}\mathcal{G}$ over $f_ a^{-1}(U_ i)$ for some $a : j \to i$. For $b : k \to j$ let $Z_ k \subset f_{a \circ b}^{-1}(U_ i)$ be the closed subset of points $x$ such that the image of $s$ and $s'$ in the stalk $(f_{a \circ b}^{-1}\mathcal{G})_ x$ are different. If $Z_ k$ is nonempty for all $b : k \to j$, then by Topology, Lemma 5.24.2 we see that $\mathop{\mathrm{lim}}\nolimits _{b : k \to j} Z_ k$ is nonempty too. Then for $x \in \mathop{\mathrm{lim}}\nolimits _{b : k \to j} Z_ k \subset X$ (observe that $\mathcal{I}/j \to \mathcal{I}$ is initial) we see that the image of $s$ and $s'$ in the stalk of $p_ i^{-1}\mathcal{G}$ at $x$ are different too since $(p_ i^{-1}\mathcal{G})_ x = (f_{b \circ a}^{-1}\mathcal{G})_{p_ k(x)}$ for all $b : k \to j$ as above. Thus if the images of $s$ and $s'$ in $p_ i^{-1}\mathcal{G}(p_ i^{-1}(U_ i))$ are the same, then $Z_ k$ is empty for some $b : k \to j$. This proves injectivity.
Surjectivity. Let $s$ be a section of $p_ i^{-1}\mathcal{G}$ over $p_ i^{-1}(U_ i)$. By Topology, Lemma 5.24.5 the set $p_ i^{-1}(U_ i)$ is a quasi-compact open of the spectral space $X$. By construction of the pullback sheaf, we can find an open covering $p_ i^{-1}(U_ i) = \bigcup _{l \in L} W_ l$, opens $V_{l, i} \subset X_ i$, sections $s_{l, i} \in \mathcal{G}(V_{l, i})$ such that $p_ i(W_ l) \subset V_{l, i}$ and $p_ i^{-1}s_{l, i}|_{W_ l} = s|_{W_ l}$. Because $X$ and $X_ i$ are spectral and $p_ i^{-1}(U_ i)$ is quasi-compact open, we may assume $L$ is finite and $W_ l$ and $V_{l, i}$ quasi-compact open for all $l$. Then we can apply Topology, Lemma 5.24.6 to find $a : j \to i$ and open covering $f_ a^{-1}(U_ i) = \bigcup _{l \in L} W_{l, j}$ by quasi-compact opens whose pullback to $X$ is the covering $p_ i^{-1}(U_ i) = \bigcup _{l \in L} W_ l$ and such that moreover $W_{l, j} \subset f_ a^{-1}(V_{l, i})$. Write $s_{l, j}$ the restriction of the pullback of $s_{l, i}$ by $f_ a$ to $W_{l, j}$. Then we see that $s_{l, j}$ and $s_{l', j}$ restrict to elements of $(f_ a^{-1}\mathcal{G})(W_{l, j} \cap W_{l', j})$ which pullback to the same element $(p_ i^{-1}\mathcal{G})(W_ l \cap W_{l'})$, namely, the restriction of $s$. Hence by injectivity, we can find $b : k \to j$ such that the sections $f_ b^{-1}s_{l, j}$ glue to a section over $f_{a \circ b}^{-1}(U_ i)$ as desired.
$\square$
Next, in addition to the cofiltered system $X_ i$ of spectral spaces, assume given
a sheaf $\mathcal{F}_ i$ on $X_ i$ for all $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$,
for $a : j \to i$ an $f_ a$-map $\varphi _ a : \mathcal{F}_ i \to \mathcal{F}_ j$
such that $\varphi _ c = \varphi _ b \circ \varphi _ a$ whenever $c = a \circ b$. Set $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits p_ i^{-1}\mathcal{F}_ i$ on $X$.
Lemma 6.29.4. In the situation described above, let $i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{I})$ and let $U_ i \subset X_ i$ be a quasi-compact open. Then
\[ \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ j(f_ a^{-1}(U_ i)) = \mathcal{F}(p_ i^{-1}(U_ i)) \]
Proof.
Recall that $p_ i^{-1}(U_ i)$ is a quasi-compact open of the spectral space $X$, see Topology, Lemma 5.24.5. Hence Lemma 6.29.1 applies and we have
\[ \mathcal{F}(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} p_ j^{-1}\mathcal{F}_ j(p_ i^{-1}(U_ i)). \]
A formal argument shows that
\[ \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathcal{F}_ j(f_ a^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_ b^{-1}\mathcal{F}_ j(f_{a \circ b}^{-1}(U_ i)) \]
Thus it suffices to show that
\[ p_ j^{-1}\mathcal{F}_ j(p_ i^{-1}(U_ i)) = \mathop{\mathrm{colim}}\nolimits _{b : k \to j} f_ b^{-1}\mathcal{F}_ j(f_{a \circ b}^{-1}(U_ i)) \]
This is Lemma 6.29.3 applied to $\mathcal{F}_ j$ and the quasi-compact open $f_ a^{-1}(U_ i)$.
$\square$
Comments (2)
Comment #8679 by Peter Fleischmann on
Comment #9387 by Stacks project on