# Wednesday, October 20

:::{.remark}
Today: only the most important property of schemes, the existence of fiber products!
Let $X, Y\in \Sch\slice S$, then the fiber product $X\fiberprod{S} Y \in \Sch\slice{S}$ is an object satisfying a universal property:

\begin{tikzcd}
	{\forall Z} \\
	\\
	&& {X\fiberprod{S} Y} && X \\
	&&& {} \\
	&& Y && Z
	\arrow[from=5-3, to=5-5]
	\arrow[from=3-5, to=5-5]
	\arrow["{\exists! \theta}"{description}, dashed, from=1-1, to=3-3]
	\arrow["\beta"{description}, from=1-1, to=3-5]
	\arrow["\alpha"{description}, from=1-1, to=5-3]
	\arrow["p"{description}, from=3-3, to=5-3]
	\arrow["q"{description}, from=3-3, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNixbMiwyLCJYXFxmcHtTfSBZIl0sWzMsM10sWzQsMiwiWCJdLFsyLDQsIlkiXSxbNCw0LCJaIl0sWzAsMCwiXFxmb3JhbGwgWiJdLFszLDRdLFsyLDRdLFs1LDAsIlxcZXhpc3RzISBcXHRoZXRhIiwxLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV0sWzUsMiwiXFxiZXRhIiwxXSxbNSwzLCJcXGFscGhhIiwxXSxbMCwzLCJwIiwxXSxbMCwyLCJxIiwxXV0=)

Note that needing the square involving $Z$ and $X\fiberprod{S} Y$ to commute is automatic, since we're working in $\Sch\slice S$ instead of just $\Sch$.
Note that $X\times Y = X\fiberprod{\spec \ZZ} Y$ recovers the product.
:::


:::{.question}
Do fiber products exist?
They're unique up to isomorphism if they do, so we just need to construct it.
:::


:::{.theorem title="Existence of fiber products"}
Fiber products exist and are unique up to isomorphism.
:::

## The 7-Step Proof

:::{.proof title="Step 1: Prove for $X, Y, S$ are affine."}
Write $X = \spec A, Y= \spec B, S = \spec R$.
We start in $\Ring$, and noting contravariance of $\spec(\wait)$, the claim is that $\spec (A\tensor_R B) \cong X\fiberprod{S} Y$.
Use that $\spec: \Ring \to \Sch$ is fully faithful, which almost allows just reversing arrows if some care is taken.
A map $Z\to \spec A$ is the same data as a map $A\to \OO_Z(Z)$.
Let $\mcu \covers Z$ with $U_i = \spec C_i$, then a morphism $Z\to \spec A \in \Sch\slice S$ is equivalently a collection of compatible morphisms $A\to C_i \in \Ring$ by the sheaf condition, so the restrictions to $\OO_{Z_i \intersect Z_j}$ are compatible.
So we can interchange any two diagrams of the following form:

\begin{tikzcd}
	Z & Y && {\OO_Z(Z)} & B \\
	X & S && A & R
	\arrow[from=1-1, to=1-2]
	\arrow[""{name=0, anchor=center, inner sep=0}, from=1-2, to=2-2]
	\arrow[from=1-1, to=2-1]
	\arrow[from=2-1, to=2-2]
	\arrow[from=2-5, to=1-5]
	\arrow[from=1-5, to=1-4]
	\arrow[from=2-5, to=2-4]
	\arrow[""{name=1, anchor=center, inner sep=0}, from=2-4, to=1-4]
	\arrow[shorten <=14pt, shorten >=14pt, Rightarrow, from=0, to=1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOCxbMCwwLCJaIl0sWzAsMSwiWCJdLFsxLDEsIlMiXSxbMSwwLCJZIl0sWzMsMCwiXFxPT19aKFopIl0sWzQsMCwiQiJdLFszLDEsIkEiXSxbNCwxLCJSIl0sWzAsM10sWzMsMl0sWzAsMV0sWzEsMl0sWzcsNV0sWzUsNF0sWzcsNl0sWzYsNF0sWzksMTUsIiIsMCx7InNob3J0ZW4iOnsic291cmNlIjoyMCwidGFyZ2V0IjoyMH19XV0=)

Now the universal property of $A\tensor_R B \in \Ring$ yields a unique map $A\tensor_R B \mapsvia{\theta*} \OO_Z(Z)$, so equivalently $Z \mapsvia{\theta} \spec(A\tensor_R B)$.
:::

For step 2, the universal property will imply uniqueness if it exists, which we'll need for gluing.

:::{.proof title="Step 3: Gluing morphisms on covers"}
A morphism $X \to Y\in \Sch$ is equivalently the data of $\mcu \covers X$ and morphism $U_i \mapsvia{f_i} Y \in \Sch$ with $\ro{f_i}{U_i \intersect U_j} = \ro{f_j}{U_i \intersect U_j}$.
This is true more generally for any $X\in \Top$ and $F\in \Sh\slice X(\cat{C})$ with values in any category.
:::


:::{.proof title="Step 4: Passing to open subsets of a factor"}
Let $X, Y\in \Sch\slice S$ be arbitrary and $U \subseteq X$ open.
If $X\fiberprod{S} Y$ exists, it is equipped with morphisms $q$ to $X$ and $p$ to $Y$.
Note that every open subset has a canonical open subscheme structure.

:::{.claim}
$U \fiberprod{S} Y \cong p\inv(U)$, noting that we don't yet know that fibers are schemes.
:::


:::{.proof title="?"}
Let $Z\in \Sch\slice S$ such that we have the following:

\begin{tikzcd}
	Z && U && X \\
	\\
	Y
	\arrow["\iota", hook, from=1-3, to=1-5]
	\arrow["\alpha", from=1-1, to=1-3]
	\arrow["\beta"', from=1-1, to=3-1]
	\arrow["{\alpha'}", curve={height=-30pt}, from=1-1, to=1-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJaIl0sWzIsMCwiVSJdLFs0LDAsIlgiXSxbMCwyLCJZIl0sWzEsMiwiXFxpb3RhIiwwLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMCwxLCJcXGFscGhhIl0sWzAsMywiXFxiZXRhIiwyXSxbMCwyLCJcXGFscGhhJyIsMCx7ImN1cnZlIjotNX1dXQ==)

If $X\fiberprod{S} Y$ exists, then $\exists ! \theta: Z\to X\fiberprod{S} Y$.
Then $\Theta(Z) \subseteq p\inv(U)$ since $p \circ \Theta = \alpha' \da \iota \circ \alpha$, so $\im(p \circ \Theta) \subseteq U$.
So $\theta:Z\to p\inv(U)$ is unique, making $p\inv(U) \cong U\fiberprod{S} Y$.
:::

So if $X\fiberprod{S}Y$ exists then $U\fiberprod{S} Y$ exists.
:::

:::{.proof title="Step 5: Gluing fiber products from an open cover"}
Let $X, Y\in \Sch\slice S$ and suppose $\mcx \covers X$ where $X_i \fiberprod{S} Y$ exists, then the claim is that $X\fiberprod{S} Y$ exists.
Define $X_{ij} \da X_i \intersect X_j$, then by step 4 $p_i\inv(X_{ij})$ is a fiber product $X_{ij}\fiberprod{S} Y$, and similarly $p_j\inv(X_{ij}) = X_{ij} \fiberprod{S} Y$.
By uniqueness in step 2, there is a unique isomorphism $\phi_{ij}: p_i\inv(X_{ij})\to p_j\inv(X_{ij})$ of fiber products.
Furthermore, the cocycle condition is satisfied since $\phi_{ik}$ is unique, so $\phi_{ij} \circ \phi_{jk} = \phi_{ik}$.
These are schemes (or more generally any ringed space), so we can glue to get *some* scheme which we'll suggestively write $X\fiberprod{S} Y$.
The claim is that this satisfies the correct universal property.
First: there are morphisms $X\fiberprod{S} Y \mapsvia{p} X$ and $X\fiberprod{S} Y \mapsvia{q} Y$.
Note that the $X\fiberprod{S} Y$ cover $X_i \fiberprod{S} Y$ and $X = \disjoint X_i/\sim$
Define the following maps:

\begin{tikzcd}
	&& {Z_i \da \alpha\inv(X_i)} && {X_i} \\
	\\
	Y && Z && X
	\arrow["\alpha", from=3-3, to=3-5]
	\arrow["\beta"', from=3-3, to=3-1]
	\arrow[hook, from=1-3, to=3-3]
	\arrow[hook, from=1-5, to=3-5]
	\arrow["{\ro{\alpha}{Z_i}}", from=1-3, to=1-5]
	\arrow["{\ro{\beta}{Z_i}}"', curve={height=30pt}, from=1-3, to=3-1]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJaIl0sWzQsMiwiWCJdLFswLDIsIlkiXSxbMiwwLCJaX2kgXFxkYSBcXGFscGhhXFxpbnYoWF9pKSJdLFs0LDAsIlhfaSJdLFswLDEsIlxcYWxwaGEiXSxbMCwyLCJcXGJldGEiLDJdLFszLDAsIiIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzQsMSwiIiwyLHsic3R5bGUiOnsidGFpbCI6eyJuYW1lIjoiaG9vayIsInNpZGUiOiJ0b3AifX19XSxbMyw0LCJcXHJve1xcYWxwaGF9e1pfaX0iXSxbMywyLCJcXHJve1xcYmV0YX17Wl9pfSIsMix7ImN1cnZlIjo1fV1d)

Then $\exists ! \Theta_i: Z_i\to X_i \fiberprod{S} Y$ where the $\Theta_i$ agree on overlaps $Z_{ij}$ as morphisms $Z_{ij} \to X_{ij}\fiberprod{S} Y$.
By step 3, these glue to a unique $\Theta: Z\to X\fiberprod{S} Y$, since the gluing is defined as $X\fiberprod{S} Y = \Disjoint_i (X_i\fiberprod{S} Y)/\phi_{ij}(p) \sim p$.
Why does it make the above diagram commute?

\begin{tikzcd}
	Z \\
	\\
	&& {X\fiberprod{S} Y} && X \\
	\\
	&& Y && S
	\arrow[from=3-5, to=5-5]
	\arrow[from=5-3, to=5-5]
	\arrow[from=3-3, to=5-3]
	\arrow[from=3-3, to=3-5]
	\arrow["\Theta"{description}, from=1-1, to=3-3]
	\arrow["\alpha", from=1-1, to=3-5]
	\arrow["\beta", from=1-1, to=5-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNSxbMiwyLCJYXFxmcHtTfSBZIl0sWzIsNCwiWSJdLFs0LDQsIlMiXSxbNCwyLCJYIl0sWzAsMCwiWiJdLFszLDJdLFsxLDJdLFswLDFdLFswLDNdLFs0LDAsIlxcVGhldGEiLDFdLFs0LDMsIlxcYWxwaGEiXSxbNCwxLCJcXGJldGEiXV0=)

This commutes because such a map is determined on an open cover, and we have commutativity in the following:

\begin{tikzcd}
	{Z_i} \\
	\\
	&& {X_i\fiberprod{S} Y} && X
	\arrow["{\Theta_i}", from=1-1, to=3-3]
	\arrow["{p_i}", from=3-3, to=3-5]
	\arrow["{\alpha_i}", from=1-1, to=3-5]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJaX2kiXSxbMiwyLCJYX2lcXGZwe1N9IFkiXSxbNCwyLCJYIl0sWzAsMSwiXFxUaGV0YV9pIl0sWzEsMiwicF9pIl0sWzAsMiwiXFxhbHBoYV9pIl1d)

So $X\fiberprod{S} Y$ exists if $X_i \fiberprod{S} Y$ exists for $\mcx \covers X$.
:::

:::{.proof title="Step 6: Affine base"}
Let $S\in \Aff\Sch$, then by step 1 $X_i \fiberprod{S} Y_j$ exists, to $X_i\fiberprod{S} Y$ exists by step 5, which implies $X\fiberprod{S} Y$ exists
:::

:::{.proof title="Step 7: Arbitrary"}
Let $X,Y,S\in \Sch\slice{S}$ be arbitrary.
Take $\mcs \covers S$, and set $X_i = p\inv(S_i)$ and $Y_i = q\inv(S_i)$.
Then $X_i \fiberprod{S_i} Y_i$ exists, and the claim is that there is an isomorphism
\[
X_i\fiberprod{S_i} Y_i \cong X\fiberprod{S} Y_i \in \Sch\slice S
.\]

Then there exist $Z\to Y_i$, and $\im(Z\to S)$ must lie in $S$.

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\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/Schemes/sections/figures}{2021-10-20_12-32.pdf_tex} };
\end{tikzpicture}

:::