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Let X be a topological space and A be an abelian group. Give A the discrete topology. For any open set U of X, Let $\cal A(U)$ be the group of all continuous aps of U into A. Thus with the usual restriction maps we obtain a sheaf $\cal A$.

So, why for every connected open set U, $\cal A(U)\cong A$ for all U? And, what is the sheafifcation of the presheaf $U \mapsto A$?

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1 Answer 1

Some hints:

For your first question: $A$ has the discrete topology, so elements of $A$ are open and closed. Then, what properties does the preimage of a single element under a continuous map $U \to A$ have? Recall the definition of connectedness. What does this imply for continuous maps $U \to A$?

For your second question, write out explicitly the definition of the sheafification functor for your example and see that the result you will get is exactly the sheaf you described in the first part of your question.

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