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 $F$.

You are correct about the interpretation of the section of a constant sheaf. By definition a constant sheaf is the sheafification of a constant presheaf, the latter being defined as "all sections over opens are equal to a same set".

Now, about "your" sheaf of sections and covering space, you should look for the definition of espace étalé associated to a preshead, because that is exactly what you call "sheaf of sections" is.

  • $\begingroup$ 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. $\endgroup$ – Danny Sep 10 at 19:38
  • $\begingroup$ 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. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 10 at 19:52
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    $\begingroup$ 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 ... $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:52
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    $\begingroup$ ... 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. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:53
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    $\begingroup$ 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. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 8:56

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