# Identity Element of a Group of Sections of a Sheaf

Serres gives the following definition of a sheaf in the paper FAC:

Let $X$ be a topological space. A sheaf of abelian groups on $X$ (or simply a sheaf ) consists of:

(a) A function $x \to \mathscr{F}_x$, giving for all $x \in X$ an abelian group $\mathscr{F}_x$,

(b) A topology on the set $\mathscr{F}$, the sum of the sets $\mathscr{F}_x$.

If $f$ is an element of $\mathscr{F}_x$, we put $\pi(f) = x$; we call the mapping of $\pi$ the projection of $\mathscr{F}$ onto $X$; the family in $\mathscr{F} \times \mathscr{F}$ consisting of pairs $(f,g)$ such that $\pi(f) = \pi(g)$ is denoted by $\mathscr{F}+\mathscr{F}$.

(I) For all $f \in \mathscr{F}$ there exist open neighborhoods $V$ of $f$ and $U$ of $\pi(f)$ such that the restriction of $\pi$ to $V$ is a homeomorphism of $V$ and $U$.(In other words, is a local homeomorphism).

(II) The mapping $f \mapsto -f$ is a continuous mapping from $\mathscr{F}$ to $\mathscr{F}$, and the mapping $(f, g) \mapsto f + g$ is a continuous mapping from $\mathscr{F}+\mathscr{F}$ to $\mathscr{F}$.

If $U$ is an open subset of $X$ then a map $s: U \to \mathscr{F}$ is called a section over $U$ if $s$ is continuous and $\pi \circ s =$ id$_U$.

It is asserted that the set of all sections over a fixed subset $U$ form an abelian group with the operation of pointwise addition. I would like to like to verify this. I see that if there is at least one section $s$ on $U$, the abelian group structure will follow from (2). But how do I know there is at least one section, say $s$, for each $U$?

I tried to prove that for a given $U$, the map that takes $x \in U$ to the identity element of the corresponding stalk, $(x,e)$ is continuous, by invoking 1, but was unsuccessful.

• I think something is missing from your definition. What are (1) and (2)? – Alex Kruckman May 8 '18 at 6:21
• Rats. You are correct. Sorry, this is the first time trying to ask from my phone. I’ll edit – Prince M May 8 '18 at 6:35
• Ok, it should be all there now. Sorry for the low quality typesetting, I just walked off an airplane and really want to get this figured out. – Prince M May 8 '18 at 6:40
• @TobiasKildetoft : if you have a functor $\mathcal{F}$, that is a sheaf according to the "usual definition", then taking $\mathcal{F}_x = \varinjlim_{U\ni x}\mathcal{F}(U)$ yields a sheaf as defined here (you have to topologise it well enough). Conversely, given a sheaf as defined here, the (functor) sheaf of sections will be a sheaf as you know them – Max May 8 '18 at 10:02
• @TobiasKildetoft: In modern terms, this is defining the "étale space" of a sheaf. There is an equivalence of categories between the category of sheaves on a space $X$ and the slice category $LH/X$, where $LH$ is the subcategory of Top consisting of all spaces, but with local homeomorphisms as the maps. – Hurkyl May 8 '18 at 10:21

## 1 Answer

Consider the mapping you gave, that is $x\mapsto (x,e_x)$ where $e_x$ is the identity element of $\mathscr{F}_x$.

You want to show that it is continuous, but continuity is a local property so you can look locally; and then invoke (I) :

Let $x\in X$ and let $V$ be a neighbourhood of $(x,e_x)\in \mathscr{F}$ and $U$ a neighbourhood of $x$ such that $\pi: V\to U$ is a homeomorphism. Let $s:U\to V$ be its converse.

Denote $V\times_X V = \{(z,y) \in V, \pi(z) = \pi(y)\}$ (this is the more common notation for $V+V$). By (II), $m:V\times_X V \to V$ defined by $m(z,y) = z-y$ is continuous.

Consider now $\varphi: V\to V\times_X V$, $y\mapsto (y,y)$ which is also continuous.

Finally, let $d: U\to V$ be defined as $m\circ \varphi \circ s$. $d$ is continuous, $\pi\circ d = id_U$ is also clear from the definitions.

Moreover, unravelling the definition yields that $d:U\to V$ is precisely $d(y)= (y,e_y)$: hence $x\mapsto (x,e_x)$ is locally continuous, hence continuous.

So this gives us a global section $X\to \mathscr{F}$ which is a neutral element in the set of sections of $X$, and clearly its restriction to any open set has the same property.

Passing remark: In the functorial definition, a sheaf of groups is a functor $O(X)^{op}\to \mathbf{Grp}$ satisfying certain "gluing conditions". But it's actually easy to see that it's the same thing as a group object in the category $\mathbf{Sh}(X)$, the category of sheaves of sets on $X$. So with this definition, the continuity of the aforementioned map is automatic, because in the definition of a group object you have a map from the terminal object to the group $G$, but this essentially means a map $X\to G$ "over $X$" (seeing $\mathbf{Sh}(X)$ as $Etale(X)$)

• This is excellent, very helpful. I am happy that my approaches at least started similar, but I didn't think to define the maps $\phi$ and $m$ and reroute through $V \times_{X} V$ so I could take advantage of property (2). I was also trying to use the topological definition of pointwise continuity. Local continuity is clearly the better approach here. – Prince M May 9 '18 at 0:11
• The idea of using $(x,y)\mapsto x-y$ is indeed a very classical one whenever topology and groups are involved : for instance it's what allows you to prove that a $T_1$ topological group is $T_2$ ! You're welcome ! – Max May 9 '18 at 6:01
• Yes, that's exactly it ! Well a constant sheaf from the functorial point of view would be $\mathcal{F}(U) = A$ (for a fixed $A$) hence the restriction maps would be the identity, hence indeed from the etale space point of view this implies that the stalks are all the same (and in fact they're $A$) . So you get that the constant sheaf would be $\displaystyle\bigsqcup_{x\in X} A$ suitably topologised – Max May 9 '18 at 22:16
• Yes I seem to have made a mistake earlier, it's not $\mathcal{F}(U)$ that is constantly $A$, it's really the stalks (hence the etale space point of view makes it easier to see why we say constant sheaf) that are constantly $A$. In particular, you are right, the (functorial) sheaf of sections associated to it takes the value $0$ when evaluated at $\emptyset$. – Max May 10 '18 at 9:10
• And as of course every restriction map sends any map to this map $\emptyset \to \mathscr{F}$, they are indeed homomorphisms. But for the purposes of this technicality, it may be interesting to note that the correct definition of a section should thus be "the corestriction of $\pi\circ s$ onto its image is $id_U$" – Max May 10 '18 at 9:12