# Isomorphisms of Affine Varieties (Tuesday, October 13)

Last time: proved that if $X, Y$ are affine varieties then there is a bijection
\[  
\correspond{\text{Morphisms} \\ f:X\to Y}
&\iff
\correspond{\text{$k\dash$algebra morphisms}\\ A(Y) \to A(X)}
\\
f & \mapsto f^*: \OO_Y(Y) \to \OO_X(X)
.\]

:::{.remark}
A morphism $f:X\to Y$ is by definition a morphism of ringed spaces where $\OO_X, \OO_Y$ are the sheaves of regular functions.
This shows $X\cong Y$ as ringed spaces iff $A(X) \cong A(Y)$ as $k\dash$algebras.
:::

## Counterexample: Isomorphisms Are Not Just Bijective Morphisms

:::{.warnings title="Isomorphisms are not necessarily bijective morphisms"}
Let $X = V(y^2 - x^3) \subset \AA^2$ and define a map
\[  
f: \AA^1 &\to X \\
t &\mapsto \tv{t^2, t^3}
,\]
This is a morphism by proposition 4.7 in [@AndreasGathmann515],
since the coordinates $t^2, t^3$ are regular functions.
Then $f$ is a bijection, since we can define a piecewise inverse
\[  
f^{-1}: X &\to \AA^1 \\
\tv{x, y} &\mapsto 
\begin{cases}
y/x & x\neq 0 \\
0 & \text{else}.
\end{cases}
\]
However, $f^{-1}$ is not a morphism, since it is not in $A(X)$ and thus not a regular function on $X$.
For instance, pulling back the function $g(t) = t$ yields 
\[  
\qty{ (f\inv)^* g} \qty{ \tv{x, y} } = 
\begin{cases}
y/x & x\neq 0\\
0 & x=y=0
\end{cases}
\quad 
\not \in A(X)
.\]
Since $f$ is a morphism, however, we can still consider the corresponding map of $k\dash$algebras:
\[  
f^*: A(X) = \frac{k[x, y]}{\gens{y^2 - x^3}} &\to A(\AA^1) = k[t] \\
x & \mapsto t^2 \\
y & \mapsto t^3
,\]
but even though $f$ is a bijective morphism, it's not an isomorphism of rings:
this can be seen from the fact that $t\not \in \im f^*$.
:::

## Categorical Products

> Review of introductory category theory.

We'll define a category $\mathrm{AffVar}_k$ whose objects are affine varieties over $k$ and morphisms in $\hom(X, Y)$ will be morphisms of ringed spaces.
There is a contravariant functor $A$ into reduced[^def_reduced]
finitely generated $k\dash$algebras which sends $X$ to $A(X)$ and sends morphisms $f:X\to Y$ to their pullbacks $f^*:A(Y) \to A(X)$.

> Review of the universal property of the product.

:::{.remark}
If we have $X,Y$ affine varieties, we take $X\cross Y$ to be the categorical product instead of the underlying product of topological spaces.
We have 
\[
A(X\cross Y) \cong A(X) \tensor_k A(Y) \cong \frac{ k[x_1, \cdots, x_n, y_1, \cdots, y_m]} { I(X) \tensor 1 + 1 \tensor I(Y) }
.\] 
This recovers the product, since we have

\begin{tikzcd}
Z \ar[dr, dotted, "{\exists! H = (f, g)}"]\ar[rrd, bend left, "f"]\ar[rdd, bend right, "g"] &                       & \\
                                                                                            & X\cross Y\ar[r]\ar[d] & X  \\
                                                                                            & Y                     & \\
\end{tikzcd}
:::

:::{.remark}
Products of spaces are sent to the tensor product of $k\dash$algebras, i.e. pullbacks are sent to pushouts.
:::

:::{.remark}
Note that the groupoid associated to a group does not have products: there can only be one element, but the outer triangles will not necessarily simultaneously commute.
:::