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Here is Rotman's definition of the skyscraper sheaf: Let $A$ be an abelian group, $X$ a topological space, and $x \in X$. Define a presheaf by $x_*A(U) = \begin{cases} A & \text{if } x \in U,\\ \{0\} & \text{otherwise.} \end{cases}$ If $U \subseteq V$, then the restriction map $\rho_U^V$ is either $1_A$ or $0$.

He then goes on to say the stalks of $x_*(A)$ are $\{0\}$ except at $(x_*A)_x$ which is $A$.

I'm going to try to show this but I am not understanding what the stalks look like. So far in the book we look at sections that are continuous maps and so $[\sigma]$ would be a germ centered at $x$ in the stalk $\mathscr F_x$ and $\tau \in [\sigma]$ occurs when there exists an open set $W$ such that $\tau \vert_W = \sigma \vert_W$ (i.e., they agree on an open neighborhood of $x$).

But now our sections are just elements of either the abelian group $A$ or the abelian group $0$. What does restriction mean on a group element? (i.e., what does it mean to say that $\tau \in [\sigma] \in (x_*(A))_y$ where $x \neq y$? Looking at $\tau, \sigma$ as maps, it means there exists an open neighborhood of $y$ such that $\tau \vert_W = \sigma \vert_W$. But we just know these are elements of an abelian group. What does restriction on these elements mean?

Any clarification would be greatly appreciated. I'm eventually going to prove his last statement about what stalks look like in this sheaf, but want a better understanding of what germs in the stalk look like.

Note: There is another question on here about skyscraper sheafs and proving the above statement, but it does not really help me understand what stalks/sections/germs look like (I think it's a different definition)

Subquestion: If $P$ a presheaf, $x \in X$, $U \ni x$ and $\sigma \in P(U)$, what is $[\sigma] \in P_x$ look like? (in terms of direct limit?) What does it mean for $\tau \in [\sigma]$? In general (i.e., where $\tau, \sigma$ are not necessarily maps that we can restrict to an open set?)

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Robert Cardona
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  • What definition of stalk are you using? The only definition I am aware of is $F_x = \varinjlim_{x \in U, \ U \ \text{open}}F(U)$ which is what is being used in the question you point to. – Jim Dec 14 '13 at 17:54
  • If $\mathscr P$ is a presheaf on a space $X$, then the stalk at $x \in X$ is $\mathscr P_x = \lim_{\rightarrow x \in U} \mathscr P(U)$. But he talks about them as germs which are equivalence classes of sections that agree on a neighborhood of $x$. Here http://en.wikipedia.org/wiki/Stalk_%28sheaf%29 they have a similar definition: "Two sections are equivalent if the restrictions of the two sections coincide on some neighborhood of $x$" – Robert Cardona Dec 14 '13 at 17:57
  • Stalks are an abstraction of germs, so you can think about them as germs to gain some intuition and there are ways in which you can say that stalks actually are germs of something, but as far as proving what a stalk is you should just apply the direct limit definition. – Jim Dec 14 '13 at 18:03
  • For the wikipedia link, a "section" is just what an element of $F(U)$ is called. It doesn't mean that it's actually defined to be a map. So in your case a "section" $s \in F(U)$ is $0$ if $x \notin U$ or is an element of $A$ if $x \in U$. – Jim Dec 14 '13 at 18:06
  • Are we assuming that $X$ is $T_1$? Otherwise I'm not sure I buy that its only nonzero stalk is at $x$. –  Dec 14 '13 at 18:14
  • @Mike I thought about this. If it was hausdorff, we could find disjoint open sets and hence, they wouldn't agree, but we were not given that. – Robert Cardona Dec 14 '13 at 18:15
  • We only need $T_1$ (an open set containing each point that doesn't contain the other). But if we don't have that, then there can be more than one nonzero stalk. –  Dec 14 '13 at 18:17
  • @Jim Ok. Thanks, this helped a bit. The problem is that Rotman appears to be doing everything in terms of germs regardless. So in terms of the direct limit: My directed system would be $((P(U)){x \in U}, (\varphi_U^V){U \subseteq V}$ restriction $)$, What are my insertion morphisms? $(\alpha_U : P(U) \to P_x)_{x \in U}$? Since I'm not looking at germs anymore, what does the projection look like? And so I'll choose an arbitrary group $G$ and maps to $G$ from $P(U)$ such that it implies $P_x = A$ or $P_x = 0$? – Robert Cardona Dec 14 '13 at 18:17
  • @Mike I don't know. I don't have a good grasp of what the sections/stalks/elements of stalks look like in this case (and in general I suppose, other than germs). That's why I've asked the question. – Robert Cardona Dec 14 '13 at 18:20
  • This isn't a sheaf of sections; (He mentions we are working with arbitrary sheaves/presheaves now). He defines the stalk as a direct limit but doesn't say what the insertion morphisms are, and since I don't know what $[\sigma] \in \lim_{\rightarrow U \ni x} P(U)$ means in general, I don't know how to construct them myself. The direct limit is an object $\lim_{\rightarrow} M_i$ and insertion morphisms $(\alpha_i : M_i \to \lim_{\rightarrow} M_i)_{i \in I}$ that satisfy specific properties that make a diagram commute correct? What would insertion morphisms in this case be? – Robert Cardona Dec 14 '13 at 18:34
  • @Robert: I suspect that Rotman is just using "germ" to mean "an element of the stalk". When "sections" are not actually sections of some map the elements of a stalk are not going to be equivalence classes of maps (germs). – Jim Dec 14 '13 at 18:34
  • I understand that. But then what do the elements of the stalk look like? How can I work with them? (It's been easy with germs, but what do I know about the elements of the stalk in general?) – Robert Cardona Dec 14 '13 at 18:36
  • I'll post an answer about how the direct limit is defined in this case. – Jim Dec 14 '13 at 18:37
  • Ok! Thanks! You've been a lot of help! – Robert Cardona Dec 14 '13 at 18:37

2 Answers2

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The stalks of $x_*A$ are $\begin{cases} (x_*A)_y =A & \text {if}\: y \in \overline {\{x\}},\\ (x_*A)_y =\{0\} & \text {if} \: y \notin \overline {\{x\}} \end{cases}$
This follows immediately from the definition of a stalk and from the fact that every neighbourhood of $y$ contains $x$ if $y \in \overline {\{x\}}$, whereas in the second case there exists a neighbourhood of $y$ not containing $x$, namely $X\setminus \overline {\{x\}}$.

Edit
As an amusing exercise inspired by Martin's comment, try to convince yourself that in the case where $\overline {\{x\}}=X$ (a common situation in algebraic geometry, and we then say that $x$ is the generic point of $X$) the sheaf $x_*A$ equals the constant sheaf with stalk $A$ and thus definitely does not look like a proud sky-scraper but rather like a depressing US housing project or British council flat or French HLM ...

Georges Elencwajg
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    I have written an answer to the question essentially because I think this thread tends to make an almost trivial result appear more complicated than it is. – Georges Elencwajg Dec 14 '13 at 19:44
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    Moreover, many people believe that the skyscraper sheaf has only one stalk, which is of course wrong for non-closed points. Good to see the general formula for the stalks, and yes it is really easy to prove. – Martin Brandenburg Dec 14 '13 at 19:48
  • I really enjoy your answer, and I am familiar with the algebro-geometric case of a generic point, but I am new to sheaves and I don't have the intuition for what these sheaves should 'look like'. Why would the constant sheaf look like a depressing US housing project? Is this because it looks the same $\textit{everywhere}$? And when I hear the name 'skyscraper' I think "tall", but is the idea not that the particular stalks of a skyscraper sheaf are neccessarily 'tall' or 'big', but when you stand back and look at the sheaf from far away, since everything around them is 0, it makes – Prince M May 11 '18 at 07:23
  • the stalks stand out a lot, like when you look at an arial shot of NYC and it looks like a ton of action concentrated in a small area, and then everything else flat outside of the area? – Prince M May 11 '18 at 07:24
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    @Prince M That's how I'm imagining skyscraper sheaves to be. – Shreya Jul 31 '20 at 02:55
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Here is what the direct limit is in your case. Let $P$ be a presheaf. We'll use the notation $s_U|_V$ for the image of $s_U \in P(U)$ under the restriction map $P(U) \to P(V)$.

Let $x \in X$ be a point and let $T$ be the set of pairs $(s_U, U)$ where $U \subseteq X$ is an open set containing $x$ and $s_U \in P(U)$. Define an equivalence relation on pairs by saying $(s_u, U) \sim (s_{U'}, U')$ if there is an open set $V \subseteq U \cap U'$ containing $x$ such that $s_U|_V = s_{U'}|_V$.

Note that if you think of $P(U)$ as actually being sections of a map then what I've written above is the definition of the equivalence relation that defines germs, but now I've written it in a way that it makes sense for any presheaf.

Define $P_x$ to be the equivalence classes $T/\sim$. In short this means that you can think of the elements of $P_x$ to be sections in some $P(U)$ and two sections are equal if and only if they eventually restrict to the same section.

Now let $P$ be the skyscraper sheaf you've defined above and let $x$ be the point at which it's defined. Then the condition that the open sets contain $x$ means $P(U) = A$ for every $U$ that appears in our equivalence relation, so $T$ consists of pairs $(a, U)$ where $U$ is open and contains $x$ and $s \in A$. Two such pairs $(a, U)$ and $(b, V)$ are equal if and only if $a$ and $b$ eventually restrict to the same place. But the restrictions are the identity on $A$ so if this is true then $a = b$. Conversely if $a = b$ then $a$ and $b$ restrict to the same element of $P(U \cap V)$ and we get that $(a, U) \sim (b, V)$. So we get that $(a, U) \sim (b, V)$ if and only if $a = b$. In other words, the map $(T/\sim) \to A$ defined by $(a, U) \mapsto a$ is a bijection.

For the other direction assume $y \neq x$. To get that $P_y = 0$ you want to show that every section eventually restricts to $0$, hence $0$ is the only equivalence class in $T/\sim$. To do this you would want to show that given any open $U$ containing $y$ there is a smaller open set $V \subseteq U$ also containing $y$ but not containing $x$. As Mike points out this is not true in all topological spaces. Rotman is probably assuming that $X$ is at least $T_1$.

Jim
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