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\begin{document}
\smallskip
\centerline{\sc Math 632, Lecture 3\\ January 14, 2004}

\section{Last time}

Recall that in classical geometrical examples of ringed spaces we have

\begin{enumerate}
	\item If $(M,\O_M)$ is a ringed space then $\O_{M,m}$ is a local ring with maximal ideal
	consisting of those functions that vanish at $m$.
	
	\item If $f:M'\rightarrow M$ is a map and $f^{\#}$ is the ``compose with $f$" map $\O_{M'}\rightarrow f_*\O_M$
	then the induced map
	$$(f^{\#})_{f(m)}:\O_{M',f(m)}\longrightarrow (f_*\O_M)_{f(m)}\longrightarrow \O_{M,m}$$
	are local maps (that is, $(f^{\#})_{f(m)}(\m_{f(m)})\subseteq \m_m$).  
\end{enumerate}

Observe that there are many examples maps of local rings which are not local maps: $\C[[t]]\hookrightarrow \C((t))$
and $\Z_{(p)}\hookrightarrow \Q$ are two examples.

\section{Locally ringed spaces}

\begin{definition}
	A {\em locally $\calC$-ringed space} is a $\calC$-ringed space $(X,\O_X)$ such that for all $x\in X$,
	the ring $\O_{X,x}$ is local (in particular nonzero).
	
	The category of locally $\calC$-ringed spaces has objects consisting of locally $\calC$-ringed spaces and has
	morphisms 
	$\varphi=(f,f^{\#}):(X,\O_X)\rightarrow (Y,\O_Y)$ such that for all $x\in X$,
	$$\varphi_{f(x)}:\O_{Y,f(x)}\longrightarrow (f_*\O_X)_{f(x)}\longrightarrow \O_{X,x}$$
	is a local map.
\end{definition}

\begin{remark} {\bf Evaluation}:  Let $(X,\O_X)$ be a locally ringed space and choose $x\in X$ and open $U\subseteq X$
containing $x$.  For $f\in \O_X(U)$ we have 
$$f\in \O_X(U)\longrightarrow \O_{X,x}\longrightarrow \O_{X,x}/\m_x=k(x).$$
The image of $f$ in $k(x)$ is denoted $f(x)$.  For example, let $(X,\O_X)$ be a real analytic manifold and let $f\in \O_X(U)$.
Suppose that $f(x)=c$ (in the usual sense of evaluating a function at a point).  Then $f(x)-c\in \m_x$ so maps to zero in
$k(x)$.  

Now suppose that $K$ is a field and $(X,\O_X)$ and $(Y,\O_Y)$ are locally ringed spaces of $K$-algebras such that 
$K\rightarrow k(x)$ and $K\rightarrow k(y)$ are isomorphisms for all $x\in X$ and $y\in Y$.  Then any morphism of ringed
spaces 
$$(f,f^{\#}):(X,\O_X)\longrightarrow (Y,\O_Y)$$ is {\em automatically} local.  Indeed, 
$$\O_{Y,f(x)}\stackrel{f^{\#}_{f(x)}}{\longrightarrow} (f_*\O_X)_{f(x)}\longrightarrow \O_{X,x} $$
is a $K$-algebra map of local $K$-algebras having residue field $K$, so it will suffice to demonstrate the
following lemma:

\begin{lemma}
	Let $K$ be a field and $\varphi:A\rightarrow B$ a $K$-algebra map of local $K$-algebras, each having residue field $K$.
	Then $\varphi$ is local.
\end{lemma}

\begin{proof}
	Choose $a\in\m_A$ and suppose that $\varphi(a)\not\in \m_B$.  Then there exists $c\in K^{\times}$
	such that under the isomorphism $K\tilde{\rightarrow} B/\m_B$ we have $c\mapsto \varphi(a)$.
	Thus, $\varphi(a)-c\in\m_B$ and since $\varphi$ is a $K$-algebra map, we have $\varphi(a-c)\in\m_B$.
	But $c\in K^{\times}$ and $a\in\m_A$ so that $a-c$ is a unit in $A$, whence $\varphi(a-c)$ is a unit in $B$.
	This is a contradiction.
\end{proof}

Let $\varphi:(X,\O_X)\rightarrow (Y,\O_Y)$ be a map of $\calC$-locally ringed spaces.  Then the map
$\varphi_x:\O_{Y,\varphi(x)}\rightarrow \O_{X,x}$ induces a unique map $k(\varphi(x))\rightarrow k(x)$:
this follows from the fact that $\varphi^{\#}(\m_{\varphi(x)})\subseteq \m_x$.  We therefore have  the commutative diagram
$$
\begin{CD}
\O_Y(U) @>\varphi^{\#}>> \O_X(\varphi^{-1}(U))\\
@VVV					@VVV\\
\O_{Y,\varphi(x)}@>>\varphi_x>\O_{X,x}\\
@VVV					@VVV\\
k(\varphi(x))@>>>k(x)
\end{CD}
$$
By chasing around an element $f\in \O_Y(U)$ in the above diagram, we see that the map $k(\varphi(x))\rightarrow k(x)$
carries $f(\varphi(x))$ to $(\varphi^{\#}(f))(x)$.

If we work with locally ringed spaces of $K$-algebras such that all residue fields are $K$ then
we have a map $\O_X\rightarrow C_{X,K}$, where $C_{X,K}$ is the sheaf of $K$-valued functions
on $X$ and for {\em any} $\varphi:(X,\O_X)\rightarrow (Y,\O_Y)$ we have the commutative diagram
$$
\begin{CD}
	\O_Y@>\varphi^{\#}>> \varphi_*\O_X\\
	@VVV			@VVV\\
	C_{Y,K}@>>{\text{old pullback}}> C_{X,K}
\end{CD}
$$
However, observe that the map $\O_X\rightarrow \C_{X,K}$ need not be injective.  As an example,
take $X=\Spec \C[T]/T^2$ and let $P=(T)$ be the unique point of $X$.
Define the structure sheaf $\O_X(X)=\C[T]/T^2$ and consider the function $T$.  Then the value of $T$
at $P$ is the image of $T$ in $O_{X,P}/\m_{P} = (\C[T]/T^2)/T=\C[T]/T$, which is 0.  Thus,
the function $T$, while nonzero, is identically zero on $X$.

\end{remark}

\begin{exercise}
	Show that for the category of $\C$-manifolds, the preceding construction identifies morphisms of
	ringed spaces of $K$-algebras with morphisms in the ``old-fashioned" sense (i.e in the sense of complex
	manifolds).
\end{exercise}

\section{Some generalities on sheaves}

We can now prove some fundamental results about sheaves.  We start with

\begin{theorem}
	Let $X$ be a topological space and $\calF,\calG$ sheaves on $X$.  Let $\varphi:\calF\rightarrow\calG$
	be a morphism of sheaves.  Then $\varphi$ is an isomorphism if and only if $\varphi_x:\calF_x\rightarrow\calG_x$
	is an isomorphism for all $x$.  
\end{theorem}

\begin{proof}
	If $\varphi$ is an isomorphism, then for all $U$ the maps $\varphi(U):\calF(U)\rightarrow\calG(U)$ are isomorphisms.
	It follows that the maps $\varphi_x:\calF_x\rightarrow \calG_x$ are isomorphisms for all $x\in X$.  (Put differently,
	the association $\varphi\rightarrow \varphi_x$ is functorial).
	
	Now suppose that $\varphi_x:\calF_x\rightarrow\calG_x$ is an isomorphism for each $x$.  Fix $U$ and let $s,t\in\calF(U)$
	be two sections over $U$ with $\varphi_U(s)=\varphi_U(t)$.  Then because of the commutative diagram
	$$
	\begin{CD}
	\calF(U)@>\varphi_U>>\calG(U)\\
	@VVV		@VVV\\
	\calF_x@>>\varphi_x>\calG_x
	\end{CD}
	$$
	we see that $\varphi_x(s_x)=\varphi_x(t_x)$.  Since $\varphi_x$ is injective, we must have $s_x=t_x$ in $\calF_x$ for all $x\in U$.
	By definition of $\calF_x$ as a direct limit, for every point $x\in U$ there exists an open set $U_x$ containing $x$ such that
	$s\big|_{U_x}=t\big|_{U_x}$ in $\calF(U_x)$.  Since the $U_x$ cover $U$ and $\calF$ is a sheaf, we have $s=t$ (by unique
	glueing).  Hence $\varphi:\calF\hookrightarrow \calG$, so in particular, for every $U$ we may consider $\calF(U)$ as a {\em subset}
	of $\calG(U)$.
	
	Now suppose $s\in \calG(U)$.  Then since $\calF_x\simeq \calG_x$, there exists a neighborhood $U_x$ of $x$
	and a section $\sigma_{U_x}\in\calF(U_x)$ such that $\varphi_{U_x}(\sigma_{U_x})=s\big|_{U_x}\in\calG(U_x)$
	(since we have some $\sigma_x\in\calF_x$ which maps to $s_x$).  Since $\varphi:\calF\rightarrow\calG$
	is injective, such $\sigma_{U_x}$ are unique.  Now observe that 
	$\sigma_{U_x}\big|_{U_x\cap U_{x'}}$ and $\sigma_{U_{x'}}\big|_{U_x\cap U_{x'}}$ both map to 
	$s\big|_{U_x\cap U_{x'}}$ in $\calG(U_x\cap U_{x'})$; since the map 
	$\varphi_{U_x\cap U_{x'}}:\calF({U_x\cap U_{x'}})\rightarrow \calG({U_x\cap U_{x'}})$ is injective,
	we conclude that $\sigma_{U_x}\big|_{U_x\cap U_{x'}}=\sigma_{U_{x'}}\big|_{U_x\cap U_{x'}}$ 
	for all $U_x,U_{x'}$.  Since $\calF$ is a sheaf, we obtain $\sigma\in\calF(U)$ such that 
	$\varphi(\sigma)\big|_{U_x}=s\big|_{U_x}$ (using the same commutative diagram as above).
	Finally, since $\calG$ is a sheaf, unique glueing holds, so $\varphi(\sigma)=s$.
\end{proof}







\end{document}