# Monday, January 31

:::{.remark}
Direct image sheaf: for $\mcf \in \Sh(X), \mcg \in \Sh(Y), f\in \Top(X, Y)$, the $f_* \in [\Sh(X), \Sh(Y)]$ is defined by $f_* \mcf(U) \da\mcf(f\inv U)$.
The inverse image functor $f\inv\in [\Sh(Y), \Sh(X)]$ is slightly more complicated.
An easy case: if $\iota: S \injects Y$ is a subspace, then it is just restriction: $(\iota\inv G)(S) \da G(S)$.

Idea for sheaf space: there are strictly horizontal neighborhoods as the homeomorphic preimages of small opens in the base.
So for $\Et_{\mcg} \mapsvia{\pi} Y$ the sheaf space of $\mcg$, define the inverse image as
\[
\Et_{\iota\inv\mcg} \da \pi\inv(S) \subseteq \Et_{\mcg}
,\]
and define a basis of sections in the following way: for $s\in \mcg(U)$, set $t(U) \da s(U) \intersect \pi\inv(S) \in \Et_{\mcg}$ to be sections of $\Et_{\iota\inv\mcg}$.
Declare these to be a basis of opens, i.e. take the subspace topology for $\pi\inv(S) \subseteq \Et_{\mcg}$ in the sheaf topology on the total space.
More generally, for $f\in \Top(X, Y)$, set
\[
\Et_{f\inv \mcg} \da \Et_\mcg \fiberprod{Y} X
.\]

The fibers are identical:

\begin{tikzcd}
	\textcolor{rgb,255:red,92;green,92;blue,214}{(\pi^*)\inv(x) } && \textcolor{rgb,255:red,92;green,92;blue,214}{\pi\inv(x)} \\
	& {\Et_{f\inv \mcg}} && {\Et_{\mcg}} \\
	\\
	& X && Y \\
	\textcolor{rgb,255:red,92;green,92;blue,214}{x} && y
	\arrow["\pi", from=2-4, to=4-4]
	\arrow["f", from=4-2, to=4-4]
	\arrow["{\pi^*}"', from=2-2, to=4-2]
	\arrow["{f^*}", from=2-2, to=2-4]
	\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-4]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-3, to=5-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, maps to, from=5-1, to=5-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, Rightarrow, dashed, no head, from=1-1, to=1-3]
	\arrow[color={rgb,255:red,92;green,92;blue,214}, dashed, from=1-1, to=5-1]
	\arrow[from=1-3, to=2-4]
	\arrow[from=1-1, to=2-2]
	\arrow[from=5-1, to=4-2]
	\arrow[from=5-3, to=4-4]
\end{tikzcd}

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

The topology on $\Et_{f\inv \mcg}$ is the coarsest topology for which $\pi^*$ and $f^*$ are continuous.
This is generated by $\qty{ f\inv(s)(f\inv U)} \intersect (\pi^*)\inv(W)$ for $W \subseteq X$ open.
Define $f\inv(s) \in f\inv\mcg(f\inv(U)) \da (f\inv U)\fiberprod{U} s(U)$.
This makes the pullback continuous both vertically and horizontally.
:::

:::{.corollary title="?"}
\[
(f\inv\mcg)_y = \mcg_{f(y)}
.\]
:::

:::{.definition title="Inverse image sheaf"}
\[
f\inv \mcg \da \qty{V\mapsto \colim_{U, V \subseteq f\inv(U)} \mcg(U)}^+
.\]
:::

:::{.remark}
How to prove this coincides with the previous definition:

- Show the stalks are isomorphic,
- Show that there is a map of presheaves $(f\inv\mcg) \to f\inv \mcg$,
- Show that the map induces an isomorphism on stalks, and lift using the universal property of sheafification.

:::

:::{.exercise title="?"}
Try to prove this by commuting limits.
:::

:::{.remark}
Recall that $K\units/\OO\units \cong \bigoplus _{x\in X} (\iota_*)_* \iota_*\inv \ul{\ZZ}$ which had stalks $\ZZ$ but was not constant -- check that the local sections differ.
:::

:::{.question}
For $S\injects Y$, does every section of $\mcg$ over $S$ extend to $Y$?
:::