# Section's local behaviour of locally constant sheaves

Let $$F$$ be a locally constant sheaf on $$X$$ and $$U$$ is an open subset and $$F|U$$ is a constant sheaf. Let $$x\in U$$, now let $$s,s'$$ be two sections from $$F(U)$$ s.t. $$s(x)=s'(x)$$, can we say $$s=s'$$ in a local neighbourhood of $$x$$?

I think there are two different understandings of "constant"

I know the sections of the constant sheaf $$A$$ over an open set $$U$$ may be interpreted as the continuous functions $$U\to A$$, where $$A$$ is given the discrete topology. If $$U$$ is connected, then these locally constant functions are constant.

I feel confused with the example below, where the sections are defined on a connected set however they are not constant, I wonder what the constant means?

If $$X$$ is locally connected, locally constant sheaves are (up to isomorphism) exactly the sheaves of sections of covering spaces $$\pi:Y\to X$$.
Such a locally constant sheaf is a constant sheaf if and only the covering $$\pi$$ is trivial.
So any non trivial covering will give you a non-constant but locally constant sheaf.
The simplest example is the sheaf of sections of the two sheeted non trivial covering $$\mathbb C^*\to \mathbb C^*:z\mapsto z^2$$ or its restriction to the unit circle $$S^1\to S^1: e^{i\theta}\mapsto e^{2i\theta}$$

Let $$X$$ be a topological space and $$F$$ be a sheave of sets on $$X$$. As the category of sets on can define stalks of $$F$$ at points of $$X$$ and by definition, the stalk $$F_x$$ of $$F$$ at an $$x\in X$$ is the inductive limit of $$F(V)$$'s indexed over all the open sets $$V$$ containing $$x$$, with order relation induced by reverse inclusion. By definition (or universal property) of the inductive limit, two sections over some open $$U$$ have the same stalk at an $$x\in U$$ if and only if they coincide on some open neighborhood $$V$$ of $$x$$ included in $$U$$. Thus the answer to you first question is yes and does not depend on $$F$$ being locally constant or any other hypothesis on $$F$$.
• why does $s(x)=s'(x)$ imply they have the same stalk? Consider the the sheaf of continuous functions on $X$, $F(U)$, then the sections are continuous functions on $U$, it can be true that two continuous agree on a point but not agree nearby. – Danny Sep 10 at 19:38
• Being equal in the stalk at $x$ does not only mean having the same value at $x$. You should reverse the definition of an inductive limit, as well as the definition (that I recalled) of the stalk of a sheaf. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 10 at 19:52
• First, I'm sorry, but you used the notation $s(x)$. I don't know what it means for you, but as your initial sheaf $F$ (the one from the beginning of your question) is any sheaf, it is not necessarily a sheaf of functions on open subsets of $X$, so we necessarily understand $s(x)$ as "the image of the section $s$ in the stalk $F_x$ of $F$ at $x$". Second, "stalk of $s$ at $x$" does not make any sense. Now, usually, you don't note $s(x)$ the image of $s$ in $F_x$, you rather note it $s_x$, for a really stupid reason : if $F$ does indeed represent a sheaf of functions, you don't want ... – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:52
• ... to mix $s_x$ (image of $s$ in the stalk $F_x$ of $F$ at $x$) with $s(x)$ which is the value of the function $s$ at $x$. They are not the same object, as $s_x$ is a germ of functions in the neighborhood of $x$ while $s(x)$ is a real number, provided the sheaf $F$ is a sheaf of real valued functions. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:53
• Finally, as I already wrote it in my answer, the stalk $F_x$ of $F$ at an $x\in X$ is the inductive limit of $F(V)$'s indexed over all the open sets $V$ containing $x$, with order relation induced by reverse inclusion. I would really advise you to reread basic definitions about inductive limits etc, it'll make you file easier. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:56