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Consider this definition of the Constant Sheaf in 'Introduction to Algebraic Geometry', Justin R. Smith:

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So we have 2 groups here: the group $A$, which is the co-domain of the continuous functions, and the group of continuous functions $U \to A$, and that second group is what is assigned to the open sets in $V$ in sheaf theory.

I have several questions here:

1) what is the group operation in that second group (it cannot be composition) ? does it matter ?

2) what is the significance of the target of the continuous functions being a group ($A$) - let alone an abelian group - for the purpose of the Constant Sheaf definition ? why can't we just have any discrete set , like {0,1,2} for instance ?

3) what is the 'Constant' in Constant Sheaf referring to ? is it because the target of the continuous functions is some discrete set ?

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    $\begingroup$ The set of continuous maps is a group under pointwise multiplication: given $f,g\colon U\to A$, set $fg\colon U\to A$ by $(fg)(u)= f(u)g(u)$. The identity is the constant map $f(u)=e_A$; the inverse of $f$ is the map $f^{-1}(u) = (f(u))^{-1}$. If you just have a discrete set, then you don’t get a group. $\endgroup$ – Arturo Magidin May 18 at 4:10
  • $\begingroup$ I get it now many thanks ! $\endgroup$ – user3203476 May 18 at 4:18

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