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\begin{document}

\title{Hartshorne's Algebraic Geometry - Section 2.1}
\author{Y.P. Lee's Class}
\maketitle

\begin{description}
  
\item[2.1.1] \emph{Let $A$ be an abelian group, and define the \textbf{constant presheaf} associated to $A$ on the topological space $X$ to be the presheaf $U \mapsto A$ for all $U \neq \emptyset$, with restriction maps the identity. Show that the constant sheaf $\mathcal{A}$ defined in the text is the sheaf associated to this presheaf.}
  
  \vspace{.15 in}
  
  \textbf{Solution by Dylan Zwick}
  
  \vspace{.1 in}
  
  If we examine the constant sheaf $\mathcal{A}$ we note that for an open set $U \subseteq X$ and a continuous map $\phi: U \rightarrow A$, with $A$ given the discrete topology, there is an obvious and natural identification between $\phi(s)$, with $s \in U$, and the corresponding element in $\mathcal{F}_{s}$, where $\mathcal{F}_{s}$ is the stalk of the constant presheaf associated to $A$ on $X$. Also, we note that for any $s \in U$ there is an open set $V \subseteq U$ defined by $\phi^{-1}(\phi(s))$\footnote{We've used the discrete topology on $A$ and the fact that $\phi$ is continuous}, and an element $\phi(s) \in A = \mathcal{F}(V)$, such that for all $q \in V$ we have that $\phi(s)_{q} = \phi(q)$, where we again use the natural identification between $\phi(s)$ and the corresponding element in $\mathcal{F}_{s}$. So, the constant sheaf defined in the text is the sheafification of the constant presheaf defined here.
  
  \vspace{.35 in}

\end{description}

\end{document}
