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\begin{document}

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

\begin{description}
  
\item[2.1.7] \emph{Let $\mathcal{F} \rightarrow \mathcal{G}$ be a morphism of sheaves.}
  
  \vspace{.1 in}
  
  \begin{enumerate}
    
  \item[(a)] \emph{Show that $im(\phi) = \mathcal{F}/ker(\phi)$.}
    
    \vspace{.15 in}
    
    \textbf{Solution by Dylan Zwick}
    
    \vspace{.1 in}
    
    First we note that if $\mathcal{F}_{1}$ and $\mathcal{G}_{1}$ are isomorphic presheaves, then they're isomorphic on stalks, and as the stalk of a presheaf is equal to the stalk of the presheaf's sheafification, we have that $\mathcal{F}_{1}^{+}$ and $\mathcal{G}_{1}^{+}$ are isomorphic on stalks. For sheaves, isomorphic on stalks is equivalent to being isomorphic, and therefore $\mathcal{F}_{1}^{+}$ is isomorphic to $\mathcal{G}_{1}^{+}$. So, if two presheaves are isomorphic, then their respective sheafifications are isomorphic as well, as of course they must be.
    
    For any open set $U$ we obviously have that the image $\phi(U)$ is isomorphic to $\mathcal{F}(U)/ker(\phi(U))$, as it's a homomorphism of abelian groups, and so the presheaf image of $\phi$ is isomorphic to the quotient presheaf $U \rightarrow \mathcal{F}(U)/ker(\phi)(U)$. As the presheafs are isomorphic, their respective sheafifications must also be isomorphic, and we have:
    
    \vspace{.1 in}

    \begin{center}
      
      $im(\phi) \cong \mathcal{F}/ker(\phi)$.
      
    \end{center}
    
    \vspace{.25 in}
    
  \item[(b)] \emph{Show that $coker(\phi) = \mathcal{G}/im(\phi)$.}
    
    \vspace{.15 in}
    
    \textbf{Solution}
    
    \vspace{.1 in}
    
    To prove this we first note that if $\mathcal{G}_{1} \subseteq \mathcal{F}_{1}$ are presheaves then for any point $p$ we have:
    
    \vspace{.1 in}
    
    \begin{center}
      
      $\ds (\mathcal{F}_{1}/\mathcal{G}_{1})^{+}_{p} = (\mathcal{F}_{1}/\mathcal{G}_{1})_{p} = ((\mathcal{F}_{1})_{p}/(\mathcal{G}_{1})_{p}) = ((\mathcal{F}_{1}^{+})_{p}/(\mathcal{G}_{1}^{+})_{p})$.
      
    \end{center}
    
    \vspace{.1 in}
    
    Therefore, as they're isomorphic on stalks, we have $(\mathcal{F}_{1}/\mathcal{G}_{1})^{+} = (\mathcal{F}_{1}^{+}/\mathcal{G}_{1}^{+})$. So, now we just note that by definition $coker(U) = \mathcal{G}(U)/\phi(U)$ and so the presheaf cokernel is $\phi$ is isomorphic to the presheaf quotient of $\mathcal{G}$ by the presheaf image of $\phi$. Therefore, we have that their respective sheafifications are isomorphic, and so:
    
    \vspace{.1 in}
    
    \begin{center}
      
      $\ds coker(\phi) \cong \mathcal{G}/im(\phi)$.
      
    \end{center}
    
  \end{enumerate}
  
  \vspace{.35 in}

\end{description}

\end{document}
