# Morphisms Glue (Thursday, October 08)

We continue the proof that morphisms glue.

:::{.proof title="part b"}
We want to show that $f^*$ sends sections of $\OO_Y$ to sections of $\OO_X$ (e.g. regular functions pullback).
Let $V\subset Y$ be open and $\phi \in \OO_Y(V)$, then
\[  
\ro{f^* \phi}{U_i \intersect f^{-1} (V)}
\qty{ \ro{f^* \phi}{U_i \intersect f^{-1} (V)} }^* \phi \in 
\OO_X(U_i f^{-1} (V))
.\]

Since pullback commutes with restriction, $f^* \phi$ is the unique $k\dash$valued function for which
\[  
\ro{f^* \phi}{U_i \intersect f^{-1} V} =
\ro{f}{U_i\intersect f^{-1} V}^* \phi
.\]
and all of the latter functions agree on overlaps $U_i \intersect U_j$.
This by unique gluing, $f^* \phi \in \OO_X(f^{-1}(V))$.
:::

## Morphisms Have Regular Components

:::{.proposition title="Morphisms between affine varieties have regular functions as components"}
Let $U\subset X$ be open in an affine variety and let $Y\subset \AA^n$ be another affine variety.
Then the morphisms $U\to Y$ of ringed spaces are the maps of the form $f = \thevector{f_1, \cdots, f_n}: U\to \AA^n$ such that $f(U) \subset Y$ and $f_i \in \OO_X(U)$ for all $i$. 
:::

:::{.proof title="$\implies$"}
Assume that $f: U\to Y$ is a morphism.
Then the coordinate functions $Y\mapsvia{y_i} \AA_1$ are regular functions, since they generate $\OO_Y(Y) = k[y_1, \cdots, y_n]/I(Y)$.
Then $f^* y_i$ is a regular function, so define $f_i \da f^* y_i$.
But then $f = \tv{f_1, \cdots, f_n}$.
:::

:::{.proof title="$\impliedby$"}
Conversely suppose $f \da \tv{f_1, \cdots, f_n}: U\to Y \subset \AA^n$ is a map such that $f_i \in \OO_U(U)$.
We want to show that $f$ is a morphism, i.e. that the pullback of every regular function is regular.
We thus need to show the following:

:::{.claim}
\envlist

1. $f$ is continuous, and
2. $f^*$ pulls back regular functions.
:::

:::{.proof title="of (1)"}
Suppose $Z$ is closed, then it suffices to show $f^{-1} (Z)$ is closed.
Then $Z = V(g_1, \cdots, g_n)$ for some $g_i \in A(Y)$.
So we can write
\[  
f^{-1} (Z) = \ts{
x\in U \st g_i(f_1(x), \cdots, f_n(x)  ) = 0\, \forall i
}
.\]
The claim is that the functions $g_i$ are regular, i.e. in $\OO_U(U)$, because the $g_i$ are polynomials in regular functions, which form a ring.
This is the common vanishing locus of $m$ regular functions on $U$. 
By lemma 3.4, the vanishing locus of a regular function is closed, so $f^{-1} (Z)$ is closed.

:::

:::{.proof title="of (2)"}
For 2, let $\phi \in \OO_Y(W)$ be a regular function on $W\subset Y$ open.
Then
\[  
f^* \phi  = \phi \circ f: f^{-1} (W) &\to K \\
x &\mapsto \phi(f_1(x), \cdots, f_n(x))
.\]
We want to show that this is a regular function.
Since the $f_i$ are regular functions, they are locally fractions, so for all $x\in f^{-1} (W)$ there is a neighborhood of $U_x\ni x$ such that (by intersecting finitely many neighborhoods) all of the $f_i$ are fractions $a_i/b_i$.
Then at a point $p = \tv{f_i(x)}$ in the image, there exists an open neighborhood $W_p$ in $W$ such that $\phi = U/V$.
But then $\phi{\tv{a_i /b_i}} = (U/V)(\tv{a_i/b_i})$, which is evaluation of a fraction of functions on fractions.
:::

:::

:::{.example}
Let $Y \da V(xy-1) \subset\AA^2$ and $U\da D(x) = \AA^1\smz \subset \AA^1$.
Note that 
\[  
A(Y)    &= {k[x, y] \over \gens{xy-1}} \\
A(\AA^1) &= k[t]
,\]
and if $f_1 \da t, f_2\da t^{-1}$, then $f_1, f_2 \in \OO_U(U)$.
So we can define a map
\[  
\tv{f_1, f_2}: U &\to Y  \\
p &\mapsto \tv{p, {1\over p} }
\]
whose image lies in $Y$.
Conversely, there is a map
\[  
V(xy - 1) &\to U = D(0) \subset  \AA^1 \\
\tv{x, y} &\mapsto x
.\]
This a morphism from $V(xy - 1)$ to $\AA^1$, since the coordinates are regular functions.
Since the image is contained in $U$, the definitions imply that this is in fact a morphism of ringed spaces.
We thus have mutually inverse maps
\[  
U       &\mapscorrespond{t\mapsto \tv{t, t^{-1} }}{x\mapsfrom \tv{x, y}} V(xy-1) \\ 
,\]
so $U\cong V(xy-1)$ as ringed spaces.
:::

:::{.slogan}
Maps of affine varieties (or their open subsets) are given by functions whose coordinates are regular.
:::

## Morphisms of Varieties on the Algebra Side


:::{.corollary title="Morphisms of varieties corresponds to $k\dash$algebra morphisms of coordinate rings"}
Let $X, Y$ be affine varieties, then there is a correspondence
\[  
\correspond{\text{Morphisms } X\to Y }
&\iff
\correspond{k\dash\text{algebra morphisms } A(Y) \to A(X)} \\
X\to Y &\mapsto A(Y) \to A(X) \\
f &\mapsto f^* \OO_Y(Y) = \OO_X(X)
.\]

Thus there is an equivalence of categories between reduced[^def_reduced]
$k\dash$algebras and affine varieties.

[^def_reduced]: 
An algebra is **reduced** iff it has no nonzero nilpotent elements.

:::

:::{.proof}
We have a map in the forward direction.
Conversely, given a $k\dash$algebra morphism $g:A(Y) \to A(X)$, we need to construct a morphism $f$ such that $f^* =  g$.
Let $Y\subset \AA^n$ with coordinate functions $y_1, \cdots, y_n$.
Then $f_i = g(y_i) \in A(X) = \OO_X(X)$.
Set $f = \tv{f_1, \cdots, f_n}$.
Then by the proposition, $f$ is a morphism to $\AA^n$.
Letting $h\in A(\AA^n)$, we have

\[
(f^*h)(x) 
&= h(f(x)) \\
&= h(\tv{f_1(x) , \cdots, f_n(x)}) \\
&= h(g(y_1), \cdots, g(y_n)) \\ 
&= g(h)(x) \qquad\text{since $g$ is an algebra morphism, $h$ is a polynomial}
\]
which follows since $f_i(x) = g(y_i)(x)$, where $g:A(Y) \to A(X)$.
So $f^*(h) = g(h)$ for all $h\in A(\AA^n)$, so the pullback of $f$ is $g$.
We now need to check that it's contained in the image.
Let $h\in I(Y)$, then $f^*(h) = g(h) = 0$ since $h = 0 \in A(Y)$.
So $\im f \subset Y$. 
Since the coordinate $f_i$ are regular, this is a morphism, and we have $f^* = g$ as desired.
:::