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I am able to verify the statements about the stalk. I want to see how the direct image of the the skyscraper sheaf can be thought of as the constant sheaf.

Observation- If $P\notin U$, then $U\cap {\{P\}}^{-}= \emptyset$ so the sections are just O which is same as the section of the skyscraper sheaf.

But if If $P\in U$ , I don't see why $i_{*}(A)(U)=A(U\cap\{P\}^{-})$ is equal to A .

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  • $\begingroup$ Is $U\cap \{P\}^{-}$ connected inside $\{P\}^{-}$? $\endgroup$
    – Babai
    Jul 2 '14 at 18:51
  • $\begingroup$ Yes it is connected, because it has a dense point, hence there can't be two disjoint opens (thus $A(U \cap \overline{P})=A$). $\endgroup$
    – user10676
    Jul 2 '14 at 19:32
  • $\begingroup$ Do u mean P is the dense point of $\overline{\{P\}}$? $\endgroup$
    – Babai
    Jul 2 '14 at 19:49
  • $\begingroup$ $P$ is also a dense point in $U \cap \overline{\{P\}}$. $\endgroup$
    – user10676
    Jul 2 '14 at 22:38
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Note that since the one-point set $\{P\}$ is irreducible, so too is its closure. And as Martin points out, any open subspace of an irreducible space is irreducible. I recommend doing Ex. I.1.6 if you haven't already.

I think the thing you want to prove in the end is that if $X$ is an irreducible topological space and $Y$ is a discrete space then any continuous map $f\colon X \to Y$ is constant. If you've gone through Chapter I then I think you know a quick proof of this already: by continuity, $f(X)$ has to be irreducible.

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  • $\begingroup$ I understand what you want to say. But then why is $U \cap \overline{\{P\}}$ irreducible in $\overline{\{P\}}$ $\endgroup$
    – Babai
    Jul 2 '14 at 21:02
  • $\begingroup$ Lemma: Every non-empty open subset of an irreducible top space is irreducible. $\endgroup$ Jul 2 '14 at 22:52

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