A sheaf is a presheaf $F$ such that for all $U$ and for all covering $\{U_i\}_{i\in I} $ of $U$, $F(U)$ is the equalizer $$ F(U) \overset{f}{\longrightarrow}\prod_{i\in I} F(U_i) {\overset{g}{\longrightarrow}\atop \underset{h}{\longrightarrow} } \prod_{i,j\in I} F(U_i\cap U_j)$$

(cf. this question for a rather explicit definition of $f,g,h$)

This can be reformulated as the gluing axioms which have a more intuitive interpretation

  • two sections $s,t\in F(U)$ are identical if their restrictions to $U_i$ coincide, i.e. $$\forall\ i\in I\quad s|_{U_i} = t|_{U_i}\ \Longrightarrow \ s=t $$

  • Given a family of sections $s_i\in F(U_i)$ such that (compatibility): $s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}$ there exist a section $s\in F(U)$ such that $s_i= s|_{U_i}$

Now if one takes as the definition of a cosheaf, a precosheaf $F$ such that $F(U)$ is the coequalizer $$ \coprod_{i,j\in I} F(U_i\cap U_j) {\overset{k}{\longrightarrow}\atop \underset{l}{\longrightarrow} } \coprod_{i\in I} F(U_i) \overset{m}{\longrightarrow} F(U)$$ what would be the gluing axioms?

First steps: I'm sure that a first condition is that all "sections" (i'm not sure about this interpretation for a precosheaf...) in $F(U)$ are the sum of sections in $F(u_i)$. However, I'm still stuck on the other condition

  • 3
    $\begingroup$ Colimits are usually harder to describe explicitly, and the description differs from category to category. It is reasonably straightforward to describe what happens for a $\mathbf{Set}$-valued functor and for a $\mathbf{Ab}$-valued functor here, though. $\endgroup$
    – Zhen Lin
    Jul 28, 2014 at 18:11

2 Answers 2


In general, a $C$-valued precosheaf $F$ is a cosheaf iff for every object $T \in C$ the presheaf $\hom(F(-),T)$ is a sheaf. This should give you an intuition for cosheafs. Instead of talking about elements of $F(U)$, which are typically morphisms into $F(U)$, better think of "coelements", which might be morphisms on $F(U)$.

If you want to consider elements anyway: A precosheaf $F$ of algebraic structures is a cosheaf iff for every open covering $U = \bigcup_i U_i$ we have that $F(U) = \coprod_i F(U_i) / \sim$, where $\sim$ is the smallest congruence relation satisfying $s^{U_i} \sim s^{U_j}$ for $s \in F(U_i \cap U_j)$. Here I denote by $s^{U_i}$ the image of $s$ in $F(U_i)$, which is then mapped into $\coprod_i F(U_i)$.

Thus, if $F$ is valued in sets, this means that every section in $F(U)$ is induced by a section in $F(U_i)$, and that two sections $s \in F(U_i)$, $t \in F(U_j)$ represent the same section in $F(U)$ iff there is a sequence of open subsets $U_i=U_{i_1},\dotsc,U_{i_n}=U_j$ in the covering and sections on the intersections $U_{i_k} \cap U_{i_{k+1}}$ which are compatible and which induce $s$ (resp. $t$) for $k=1$ (resp. $k=n-1$).

If $F$ is valued in abelian groups, then every section in $F(U)$ has the form $\sum_i s_i^{U}$ for sections $s_i \in F(U_i)$ (almost all zero). In order to decide equality, it suffices to decide when $\sum_i s_i^{U}=0$. This happens iff there are sections $s_{ij} \in F(U_i \cap U_j)$ (almost all zero) such that $s_i = \sum_j s_{ij}^{U_i} - s_{ji}^{U_i}$.

  • $\begingroup$ In the 3rd paragraph, one need only introduce $U_i$ and $U_j$. The maps $k,l: \coprod_{i,j\in I} F(U_i\cap U_j) \rightarrow \coprod_{i\in I} F(U_i) $ in the coequalizer condition are determined ( by the universal property of a coproduct) by maps $ext_{U_i\cap U_j\ U_i}$ or $ext_{U_i\cap U_j\ U_j}$. There is no map $ext_{U_i\cap U_j\ U_k}$ for $k\neq i,j$ $\endgroup$
    – Noix07
    Apr 5, 2015 at 14:16
  • $\begingroup$ @user39158: You have to take the transitive closure of the relation. What I use is the general construction of coequalizers of sets (or general algebraic structures). $\endgroup$ Jul 1, 2015 at 8:28
  • $\begingroup$ I think then that in the 2nd line of that 3rd §, one should have "$s,t\in F(U)$ are equivalent if ...", but at the moment it is written "two sections $s\in F(U_i), t \in F(U_j)$ ...". And (I still have to think) for the very last equality, i tend to think that $s_{ij}^{U_i}$ and $s_{ij}^{U_j}$ should be equivalent rather than $s_{ij}^{U_i}$ and $s_{ji}^{U_i}$ $\endgroup$
    – Noix07
    Jul 1, 2015 at 8:40
  • $\begingroup$ "closure of the relation": cf. en.wikipedia.org/wiki/… and en.wikipedia.org/wiki/… $\endgroup$
    – Noix07
    Jul 7, 2015 at 13:17

The interpretation of the gluing axioms for sheaves is intimately related to the example in which $F(U)$ is some set of functions on $U$.

In order to find an analogue interpretation for cosheaves, one can either assume that a cosheaf is of this kind (claim that compactly supported functions yield a cosheaf in this answer ) and examine what it means:

  • assume that the maps $ F(U_i) \rightarrow F(U)$ in the definition of a precosheaf are extensions of the "sections" with "support" in $U_i$, denote them by $ext_{U_i U}$ (precosheaf implies in particular: $$ ext_{U_i U}\circ ext_{U_i U_i\cap U_j} = ext_{U_j U}\circ ext_{U_j U_i\cap U_j}$$
  • the arrow $m$ in the coequalizer diagram is defined by those $ext_{U_i U}$, if there were no coequalizer condition it would mean that $F(U)$ is really the coproduct of the $ \{ F(U_i)\}$ (and this should hold for all covering $ \{ U_i\}$ of $U$.): -if one considers that $F(U)$ is only a set, then it is the disjoint union of sections of the $F(U_i)$, if $F(U)$ is considered as a vector space, then it is the direct sum, if $F(U)$ is considered as an unital commutative algebra, then it is the tensor product
  • the coequalizer condition says that $m$ is the projection on the quotient spaces detailed in Martin's answer.

One can see that the disjoint union of "sections with support" in $U_i$ (even quotiented by some equivalence relation) will never give all sections in $F(U)$; Similarly the tensor product will not yield a function of one variable. The direct sum seems to work if $s \oplus t$ is identified with the pointwise sum of two functions.

The mechanical adaptation of this interpretation of the "gluing axioms" is that $m$ induces an isomorphisms $\tilde{m}: \coprod_{i\in I} F(U_i)/\sim\ \longrightarrow F(U)$ (such that $m=\tilde{m}\circ \pi_{\sim}$, projection on the quotient). The interpretation now is not gluing, but something like "summing" (coproduct, direct sum in the good example) sections with support on small regions up to equivalence gives sections of bigger regions. (in any case, sheaves and cosheaves impose relations on sections on $ \{ U_i\}$ and those of $U$, one is gluing, the other is "sum up to equivalence")

Alternatively, one can find a good prototypical example: inspired by the first § of Martin's answer let's show that the functor $ U \rightarrow \mathcal{D}'(U)$ that to an open region associates distributions with support inside it, is a cosheaf:

  • a distribution with support in $U_i \subset U$ is naturally a distribution with support in $U$ so there is a natural precosheaf structure.
  • $\mathcal{D}'(U)$ is obtained by direct sum (cosheaf with value in the category of vector spaces) of the $\mathcal{D}'(U_i)$: multiply the distribution with functions from a partition of unity. The quotient one needs to take is actually so "automatic" that I had quite some diffculty pointing it out: a distribution with support in $U_i\cap U_j$ considered first as a distribution in $U_i$ then in $U$ is the same as a the same distribution considered first as a distribution in $U_j$m then in $U$. This is not automatic in $\bigoplus_i \mathcal{D}'(U_i)$

Remark: It seems that a coequalizer is always obtained as a quotient of the coproduct w.r.t. some relation, cf. wikipedia or this question.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .