0

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.

Community
  • 1
Prince M
  • 3,893
  • 1
    I think something is missing from your definition. What are (1) and (2)? – Alex Kruckman May 08 '18 at 06:21
  • Rats. You are correct. Sorry, this is the first time trying to ask from my phone. I’ll edit – Prince M May 08 '18 at 06:35
  • 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 08 '18 at 06:40
  • You did not define "sections on $U$" - just asking because you define scheaves somewhat unusually (without using functors, restrictions, exactness, ...) – Hagen von Eitzen May 08 '18 at 06:41
  • Where is this definition from? I find it very hard to turn it into the usual definition, which assigns a group to each open set and which does not force a topology on the sum of the groups. It seems like this definition starts with the stalks? – Tobias Kildetoft May 08 '18 at 06:44
  • This is the definition from Serre’s original paper FAC. It is ‘equivalent’ to the functional definition but not immediately obvious. With some work the equivalence is clear. But, for a section ‘s’ over an open subset U, I mean a continuous map s from U to F such that the projection composed with s is the identity on U. Then the set of all sections forms on abelian group by pointwise addition – Prince M May 08 '18 at 06:49
  • @TobiasKildetoft yes it starts with stalks. – Prince M May 08 '18 at 06:51
  • 1
    @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 – Maxime Ramzi May 08 '18 at 10:02
  • 1
    @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. –  May 08 '18 at 10:21

1 Answers1

1

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 $Etale(X)$)

Maxime Ramzi
  • 43,598
  • 3
  • 29
  • 104
  • 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 09 '18 at 00:11
  • Showing the property for the global section was also helpful. I was trying to work with an arbitrary $U$. Thanks again for your help. – Prince M May 09 '18 at 00:12
  • 1
    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 ! – Maxime Ramzi May 09 '18 at 06:01
  • Can I trouble you for one more question... on this same page in FAC, gives a constant sheaf as an example of a sheaf, but does not define a constant sheaf. There appears to be some controversy on the internet about what a constant sheaf is defined to be. Using the definition of a sheaf above, should I take the definition of a constant sheaf to be a sheaf who's stalks are all the same? It seems using the categorical definition of a constant sheaf is to require certain properties for sections, which leads to equal stalks. – Prince M May 09 '18 at 20:18
  • Also, what is the slick way to justify that $m$ is continuous using axiom (2)? I tried to show that $(f,g) \mapsto (f,-g)$ was continuous, since then I could compose with the map $(f,g) \mapsto f+g$ but I cant seem to get the first map – Prince M May 09 '18 at 21:50
  • Oh wait, I think I see it. Start with $\mathscr{F} \times \mathscr{F}$, the projection map onto each component is continuous. I can compose the second projection with $g \mapsto -g$ continuous by axiom (2). But then I have $(f,g) \mapsto f$ continuous and $(f,g) \mapsto -g$ continuous as maps from $\mathscr{F} \times \mathscr{F} \to \mathscr{F}$ thus I have the map $(f,g) \to (f,-g)$ is continuous in the product topology, but then I can restrict both the domain and codomain to $\mathscr{F}+\mathscr{F}$? – Prince M May 09 '18 at 22:00
  • As you can tell my point set topology is not strong... – Prince M May 09 '18 at 22:00
  • 1
    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 – Maxime Ramzi May 09 '18 at 22:16
  • Ok, and a technicality - what is the group associated to the set of sections over the empty set? Or does $\Gamma(U,\mathscr{F})$ being an abelian group only when $U$ is a non empty subset of $X$? – Prince M May 10 '18 at 00:09
  • Er.. wait, for the empty set there would be exactly one function, the empty function, so its automatically the trivial group? – Prince M May 10 '18 at 00:10
  • 1
    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 $\emptyset$. – Maxime Ramzi May 10 '18 at 09:10
  • 1
    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 $id_U$" – Maxime Ramzi May 10 '18 at 09:12
  • In the following section, ‘Construction of Sheaves’, Serre makes some propositions about certain sheaves being isomorphic, but has yet to define a morphism of sheaves. Using the functorial definition, since each sheaf is a functor, the isomorphism would be a natural transformation. But using the definition Serre gives, how should I interpret isomorphism of sheaves? My best guess is homeomorphism that induces isomorphism on stalks? – Prince M May 11 '18 at 01:24
  • Well to define isomorphism you always only need three things : defining morphisms, composition, and identity morphisms. The last one is clear and the second one too (because we sort of know that morphisms will be maps over $X$ - the problem is which maps ?). Now a morphism is simply a map over $X$ that induces a morphism on each stalk. An isomorphism is then, as always, a morphism with an inverse (both left and right). It's a good exercise to see what this means in terms of stalks, yes (and it's probably what you just said - I'm just giving you a recipe for determining what isomorphisms are) – Maxime Ramzi May 11 '18 at 06:07
  • Should it be a map over $X$ or a map over $\mathscr{F}$, it seems if we had a fixed topological space $X$ and two sheaves $\mathscr{F}$, $\mathscr{G}$ on $X$ would we want the morphism of sheaves to be between $\mathscr{F}$ and $\mathscr{G}$? – Prince M May 11 '18 at 07:00
  • Yes it's between those two but over $X$, i.e. the obvious diagram must commute (my "over" is from french terminology "au-dessus de" I don't know whether that's english terminology as well) – Maxime Ramzi May 11 '18 at 07:22
  • Ok, so just be sure, if $g$ is my homeomorphism of sheaves that induces isomorphism on stalks, then the final commutativity requirement would be that $g$ commutes with the projections from the sheaves to $X$? – Prince M May 14 '18 at 23:24
  • Yes, that's it ! I don't know what the correct english translation for "au dessus de $X$" should be, but when you spell it out explicitly that's what it means – Maxime Ramzi May 15 '18 at 08:28
  • Great, Thanks :) – Prince M May 15 '18 at 18:14