# Ringed Spaces (Ch. 4, Tuesday, October 06)

:::{.definition title="Ringed Spaces"}
A **ringed space** is a topological space $X$ together with a sheaf $\OO_X$ of rings.
:::

:::{.example}
\envlist

1. $X$ an affine variety and $\OO_X$ its ring of regular functions.

2. $X$ a manifold over $\RR^n$ with $\OO_X$ a ring of smooth or continuous functions on $X$.

3. $X = \ts{p, q}$ with the discrete topology and $\OO_X$ given by $p\mapsto R, q\mapsto S$.

4. Let $U\subset X$ an open subset of $X$ an affine variety.
  Then declare $\OO_U$ to be $\ro{\OO_X}{U}$.
:::

Recall that the restriction of a sheaf $\mathcal{F}$ to an open subset $U\subset X$ is defined by $\ro{\mathcal{F}}{U}(V) = \mathcal{F}(V)$.

:::{.example title="Skyscraper Sheaf"}
Let $X$ be a topological space and $p\in X$ a point.
The **skyscraper sheaf at $p$** is defined by
\[  
k_p(U) \da 
\begin{cases}
k & p\in U \\
0 & p\not\in U
\end{cases}
.\]
:::

:::{.remark}
As a convention, we'll always assume that $\OO_X$ is a sheaf of functions, so $\OO_X(U)$ is a subring of all $k\dash$valued functions on $U$.
Moreover, $\res_{UV}$ is restriction of $k\dash$valued functions.
:::

## Morphisms of Ringed Spaces

:::{.definition title="Morphisms of Ringed Spaces"}
A **morphism of ringed spaces**
\[  
(X, \OO_X) \mapsvia{f}  (Y, \OO_Y)
\]
is a continuous map $X\to Y$ such that for all opens $U \subset Y$ and any $\phi \in \OO_Y(U)$, the pullback satisfies $f^* \phi \in \OO_X(f\inv(U))$.
:::

:::{.slogan}
Pullbacks of regular functions are regular.
:::

:::{.remark}
We'll need to use th convention that $\OO_X$ is a sheaf of $K\dash$valued functions in order to make sense of pullbacks. 
In general, for schemes with $U \subset Y$ and $f^{-1} (U) \subset X$, we'll need some analog of $f^*: \OO_Y(U) \to \OO_X(f^{-1} (U) )$ to make sense of "composing" or "restricting" sections.
We still need continuity, however, so that $f^{-1}(U)$ is open when $U$ is open and thus $\OO_X(f^{-1}(U))$ makes sense.
:::

:::{.example}
If $(X, \OO_X)$ is a ringed space associated to an affine variety, then we assume $\OO_X(U)$ are literally functions on $U$.
Morphisms of open subsets is again defined by morphisms of ringed spaces.
:::

:::{.example}
Let $X = \AA^1/k$ and $U \da D(x)$, then $D(f) = \AA^1\smz$.
Then $\iota: U\injects X$ is continuous.
Given an arbitrary distinguished open set $D(f) \subset \AA^1$, we know from previous results that
\[  
\OO_X(D(f)) \da \OO_{\AA^1}(D(f)) = A(\AA^1)_{\gens{f}} = k[x]_{\gens{f}} \da \ts{g/f^n \st g\in k[x]}
.\]
We want to show that $\iota: (U, \OO_U) \injects (X, \OO_X)$ is a morphism of ringed spaces where $\OO_U(V) \da \OO_X(V)$.
Does $\iota^*$ pull back regular functions to regular functions?
Yes, since 
\[
\iota^{-1} (D(f)) = D(f) \union D(x) = D(xf)
\]
and thus 
\[
g/f^n \in \OO_U(\iota^{-1}(D(f))) = \OO_U(D(xf))
\]
where we've used that $f^n \neq 0 \implies xf\neq 0$.
:::

:::{.example}
A non-example: take
\[  
h: \AA^1 &\to \AA^1 \\
x & \mapsto 
\begin{cases}
x & x \neq \pm 1 \\
-x & x= \pm 1
\end{cases}
.\]
This is continuous because the Zariski topology on $\AA^1$ is the cofinite topology (since the closed sets are finite), so any injective map is continuous since inverse images of cofinite sets are again cofinite.


:::{.question}
Does $h$ define a morphism of ringed spaces?
I.e., is the pullback of a regular function on an open still regular?
:::

:::{.answer}
Take $U = \AA^1$ and the regular function $x\in \OO_{\AA^1}(\AA^1)$.
Then $h^*x = x\circ h$, so
\[  
(x\circ h)(p) =
\begin{cases}
p & p\neq \pm 1 \\
-p & p= \pm 1 
\end{cases}
\not \in k[x]
\]
since this is clearly not a polynomial: if two polynomials agree on an infinite set of points, they are equal.
:::

:::

:::{.example}
Consider 
\[
\iota: (\RR^2, C^\infty) \injects (\RR^3, C^\infty)
\]
is the inclusion of a coordinate hyperplane.
To say that this is a morphism of ringed spaces, we need that for all $U\subset \RR^3$ open and $f:U\to \RR$ a smooth function, we want $i^* f\in C^\infty (\iota^{-1}(U))$.
But this is the same as $f\circ \iota \in C^\infty(\RR^2\intersect U)$, which is true.
:::

## Gluing Morphisms

:::{.proposition title="Ringed spaces form a category"}
\envlist

1. They can be composed: if $\phi \in \OO_Z(U)$, then $g^* \phi \in \OO_Y(g^{-1}(U))$ and so $f^* g^* \phi \in \OO_X(f^{-1} g^{-1} (U))$.

2. The identity is a morphism.

Thus ringed spaces form a category, since composition is associative.
:::



:::{.lemma title="Gluing for Morphisms"}
Let $f:X\to Y$ be a continuous map between ringed spaces.
Assume there exists an open cover $\ts{U_i}_{i\in I}\covers X$ such that $\ro{f}{U_i}$ is a morphism, then $f$ is a morphism.

:::

:::{.remark}
Slogan: it suffices to check a morphism on an open cover.
:::

:::{.proof title="of part (a)"}
Part a:
Need to check that $f$ is continuous, can compute
\[  
f^{-1}(V) = \Union_{i\in I} U_i \intersect f^{-1}(V) = \Union_{i\in I} \ro{f}{U_i}^{-1} (V)
.\]
but the latter is open as a union of open sets, where each constituent set is open by assumption.
:::