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Many times you have to show that a certain sheaf, maybe obtained as gluing of other sheaves, is not constant ; there are methods or tricks immediate or generally to do this? What is special about a constant sheaf that other sheaves (e.g. those locally constant ) don't have ?

in particular cases :

  • if $X$ is not connected and disjoint union of two open not empty subset $U, V$ and if i take the sheaves constant on $U$ and $V$ and glue them (for example i can take the zero sheaf on $U$ and the constant sheaf with stalk $K$ on $V$); why I don't get a constant sheaf?

  • if $X = \mathbb{R}\cup S^{1}$, why the sheaf locally constant that I get as gluing of the two constant sheaves $\mathbb{R}_{\mathbb{R}}$ and $\mathbb{R}_{\mathbb{S^{1}}}$ is not constant ?.

The Number Theorist
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  • Possible duplicate of http://math.stackexchange.com/questions/606952/locally-constant-sheaf-but-not-constant – Peter Franek Jan 26 '16 at 17:03
  • I already read the question that you linked me; But i would have more details in this two example and i would understand in general how approach this problem. – The Number Theorist Jan 26 '16 at 17:07
  • But isn't it answered there? If the underlying space is the circle $S^1$ and the sheaf consists of sections of the map $z\mapsto z^2$, then the sheaf locally assigns to a set $U$ the 2-element group ${\pm \sqrt{z}}$ (considered as two functions on $U$). But the square root can not be extended to a global function. – Peter Franek Jan 26 '16 at 17:11
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    of course! i understood this example perfectly . But I don't know how to carry these reasoning in my two problems ; for example, the first problem that I have written seems to need a very different approach respect to the example you're talking about. – The Number Theorist Jan 26 '16 at 17:22

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