# Hartshorne Exersice 1.17 Skyscraper sheaf Chapter II Schemes

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 .

• Is $U\cap \{P\}^{-}$ connected inside $\{P\}^{-}$? Jul 2 '14 at 18:51
• Yes it is connected, because it has a dense point, hence there can't be two disjoint opens (thus $A(U \cap \overline{P})=A$). Jul 2 '14 at 19:32
• Do u mean P is the dense point of $\overline{\{P\}}$? Jul 2 '14 at 19:49
• $P$ is also a dense point in $U \cap \overline{\{P\}}$. Jul 2 '14 at 22:38

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.
• I understand what you want to say. But then why is $U \cap \overline{\{P\}}$ irreducible in $\overline{\{P\}}$ Jul 2 '14 at 21:02