Given a cover $\{U_i\}$ of a space $X$ and for each $U_i$ a sheaf $\mathcal{F}_i$ and isomorphisms $\phi_{ij}:\mathcal{F_j}|_{U_i \cap U_j} \rightarrow \mathcal{F_i}|_{U_i \cap U_j}$ satisfying the cocycle condition $\phi_{ij}\phi_{jk}\phi_{ki} = id$, I want to show that there is a sheaf $\mathcal{F}$ on $X$ whose restriction to each $U_i$ is isomorphic to $\mathcal{F_i}$. I understand that $\mathcal{F}(U)$ should consist of tuples $(s_i)$ of sections from the various sheaves, then I need to define restriction maps, prove the presheaf axioms and then the sheaf axioms. I have at least a vague idea of how the proof should go. Unfortunately, on my way to proving the result I get utterly lost in an unholy mess of details, so that even once I "finished" my proof I was nowhere near certain my proof was correct. I've searched around for proofs to examine, but any proof I can find is either lacking many details or required as an exercise.

Here are my questions:

1) Can someone show me a proof of this result? I am interested to see both a direct proof checking all the details and also a more intuitive proof, possibly appealing to results about gluing morphisms or whatnot.

2) Should the tuples $(s_i)$ be leaving in the product or the disjoint union of the $\mathcal{F_i}(U\cap U_i)$? I thought it would be product, but I saw union on Google Books in Introduction to Singularities and Deformations by Greuel, Lossen, and Shustin.

3) Where does the cocycle condition come into play?

4) How can I prevent all these messy sheaves from deterring me from the beautiful subject of algebraic geometry?

Thanks in advance.

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    $\begingroup$ Two links I found helpful: stacks.math.columbia.edu/tag/00AK & stacks.math.columbia.edu/tag/01JJ $\endgroup$ Commented Aug 12, 2014 at 15:00
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    $\begingroup$ Your first paragraph sounds like any typical software development endeavour :) ! $\endgroup$ Commented Jun 6, 2016 at 14:23
  • $\begingroup$ As have been discussed, The cocycle condition is not needed to make this construction work. Another point why the cocycle condition is needed other than the answer below is that to glue the topological space, the cocycle condition is used to guarantee the transitive of the equivalence relation see math.stackexchange.com/a/3852875/360262 $\endgroup$
    – yi li
    Commented Sep 29, 2022 at 2:11

1 Answer 1


You should really be able to do this directly. The calculations are not messy at all, in my opinion. Also, they are straight forward. It is also a good idea to simplify the notation and to use words more than formulas. This way you really understand what is going on.

You define $F(U)$ to be the set of all families $s=(s_i)$ of sections $s_i \in F_i(U \cap U_i)$ which are compatible in the sense that $\phi_{ij}(s_j)=s_i$ for all $i,j$. I have simplified the notation here: Of course we restrict $s_i$ and $s_j$ to $U \cap U_i \cap U_j$, and of course we apply $\phi_{ij}$ at this open subset.

The restriction maps of $F$ are induced by the ones for $F_i$. They are well-defined because the asserted compatibility is preserved by restriction, which in turn works since $\phi_{ij}$ commutes with restriction maps. After having checked this, it is obvious that $F$ becomes a presheaf, using the presheaf properties of the $F_i$.

Now as for the first sheaf condition, let $s=(s_i)$ be as above, and $U = \cup_p W_p$ be an open cover. If $s$ is trivial on each $W_p$, this means that $s_i \in F_i(U \cap U_i)$ is trivial on each $W_p \cap U_i$. But since these cover $U \cap U_i$, it follows $s_i=0$, for all $i$, hence $s=0$. (For sheaves of sets, you can adjust this argument easily.)

For the second sheaf condition, let $s^p \in F(W_p)$ be a family of compatible sections (compatiblity means that $s^p$ and $s^q$ agree on $W_p \cap W_q$). This means that for every $i$ we have a family of compatible sections $s^p_i \in F_i(W_p \cap U_i)$ with respect to the cover $\{W_p \cap U_i\}$ of $W \cap U_i$. Since $F_i$ is a sheaf, these glue to a section $s_i \in F_i(W \cap U_i)$. We have $\phi_{ij}(s_j)=s_i$ in $F_i(U \cap U_i \cap U_j)$, since this is true when restricted to each $W_p \cap U_i \cap U_j$, since $s^p \in F(W_p)$. Hence, $s:=(s_i) \in F(U)$ and $s$ restricts to $s^p$ on $W_p$ by construction.

Thus, $F$ is a sheaf.

The cocycle condition is not needed to make this construction work. We don't even need that the $\phi_{ij}$ are isomorphisms! This is especially clear in the category-theoretic construction of $F$, see for example Zhen Lin's answer here.

But there is a reason why one usually demands this condition: We would like to have that the projection $F|_{U_i} \to F_i$ mapping a section $s$ to $s_i$ is an isomorphism. We simply construct an inverse by mapping $s_i$ to $s$ defined by $s_j = \phi_{ij}^{-1}(s_i)$ (here we need that $\phi_{ij}$ is an isomorphism). This is consistent when $\phi_{ii}=\mathrm{id}$ (which would follow from the cocycle condition). By construction $\phi_{ij}(s_j)=s_i$, but in order to be a section of $F$, we also need $\phi_{kj}(s_j)=s_k$ for all $k$, i.e. $\phi_{kj} = \phi_{ki} \circ \phi_{ij}$, which is precisely the cocycle condition. One then checks that this describes a map $F_i \to F|_{U_i}$ which is inverse to the projection.

There is even an a priori motivation for the cocycle condition. Given a gluing datum of sheaves $(F_i,\phi_{ij})$, we want to find a sheaf $F$ with isomorphisms $F|_{U_i} \cong F_i$, but in such a way that the induced isomorphisms $F_j|_{U_i \cap U_j} \cong F|_{U_i \cap U_j} \cong F_i|_{U_i \cap U_j}$ really equal $\phi_{ij}$. But these isomorphisms obviously satisfy the cocycle condition: If we compose (let me again simplify the notation) $F_k \to F \to F_j$ with $F_j \to F \to F_i$, then $F \to F_j \to F$ cancels to the identity, so that we get $F_k \to F \to F_i$. In other words, in the following diagram, the outer triangle commutes because all three inner triangles commute:


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    $\begingroup$ +1 for: "The cocycle condition is not needed to make this construction work." $\endgroup$ Commented May 17, 2014 at 11:14
  • $\begingroup$ There is another way of doing this via first defining a sheaf on the base consisting of all open sets contained in some $U_i$. It is sketched out here: stacks.math.columbia.edu/tag/00AK. $\endgroup$
    – user5826
    Commented May 9, 2021 at 19:24
  • $\begingroup$ However, I can't see how the cocycle condition is used there. Any thoughts? $\endgroup$
    – user5826
    Commented May 9, 2021 at 19:26
  • $\begingroup$ @user46372819 Read the last paragraphs of my answer. $\endgroup$ Commented May 9, 2021 at 22:19
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    $\begingroup$ And comparing your answer with Zhen Lin’s, It seems that Zhen Lin’s construction does not need to use the cocycle condition as well. But he/she claimed that the cocycle condition is needed to verify that the triangles commute. I cannot see how the condition is really used there and I don’t think that this is needed. Am I misunderstood anything? Thank you for your help! :) $\endgroup$
    – Hetong Xu
    Commented May 10, 2021 at 2:33

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