# Passing to stalks, pushforward/inverse image (Friday, August 27)

## Isomorphism $\iff$ isomorphism on stalks

:::{.theorem title="Sheaf isomorphism $\iff$ isomorphism on stalks"}
Let $\phi:\mcf\to\mcg$ be a morphism in $\Sh(X)$, then $\phi$ is an isomorphism $\iff$ $\phi_p: \mcf_p \to\mcg_p$ is an isomorphism for all $p\in X$.
:::

:::{.proof title="$\implies$"}
Suppose $\phi$ is an isomorphism, so there exists a $\psi: \mcg\to \mcf$ which is a two-sided inverse for $\phi$.
Then $\psi_p$ is a two-sided inverse to $\phi_p$, making it an isomorphism.
This follows directly from the formula:
\[
\phi_p: \mcf_p &\to \mcg_p \\
(s, U) & \mapsto (\phi(U)(s), U)
.\]

:::

:::{.proof title="$\impliedby$"}
It suffices to show $\phi(U): \mcf(U) \to \mcg(U)$ is an isomorphism for all $U$.
This is because we could define $\psi(U):\mcg(U) \to \mcf(U)$ and set $\phi\inv(U) \da \psi(U)$, then reversing the arrows in the diagram for a sheaf morphism again yields a commutative diagram.


:::{.claim}
$\phi(U)$ is injective.
:::
For $s\in \mcf(U)$, we want to show $\phi(U)(s) = 0$ implies $s=0$.
Consider the germs $(s, U) \in \mcf_p$ for $p\in U$, we have $\phi_p(s, U) = (0, U) = 0\in \mcf_p$.
So $S_p = 0$ for all $p\in U$.
Since we have a germ, there exists $V_p \ni p$ open such that $\ro{s}{V_p} = 0$.
Noting that $\ts{V_p \st p\in U}\covers U$, by unique gluing we get an $s$ where $\ro{s}{V_p} = 0$ for all $V_p$, so $s\equiv 0$ on $U$.


:::{.claim}
$\phi(U)$ is surjective.
:::

Let $t\in \mcg(U)$, and consider germs $t_p\in \mcg_p$.
There exists a unique $s_p\in \mcf_p$ with $\phi_p(s_p) = t_p$, since $\phi_p$ is an isomorphism of stalks by assumption.
Use that $s_p$ is a germ to get an equivalence class $(s_p, V)$ where $V \subseteq U$.
We have $\phi(V)(s(p), V) \sim (t, U)$, noting that $s$ depends on $p$.
Having equivalent germs means there exists a $W(p) \subseteq V$ with $p\in W$ with $\phi(W(p)) \qty{\ro{s(p)}{W}} = \ro{t}{W(p)}$.
We want to glue these $\ts{ \ro{s(p)}{W(p)} \st p\in U }$ together.
It suffices to show they agree on intersections.
Taking $p, q\in U$, both $\ro{s(p)}{W(p) \intersect W(q)}$ and $\ro{s(q)}{W(p) \intersect W(q)}$ map to $\ro{t}{W(p) \intersect W(q)}$ under $\phi(W(p) \intersect W(q) )$.
Injectivity will force these to be equal, so $\exists ! s \in \mcf(U)$ with $\ro{s}{W(p)} = s(p)$.
We want to now show that $\phi(U)(s) = t$.
Using commutativity of the square, we have $\phi(U)(s) \ro{}{W(p)} = \phi(W(p)) \qty{\ro{s}{W(p)} }$.
This equals $\phi(W(p))(s(p)) = \ro{t}{W(p)}$.
Therefore $\phi(U)(s)$ and $t$ restrict to sections $\ts{w(p) \st p\in U}$.
Using unique gluing for $\mcg$ we get $\phi(U)(s) = t$.
:::

:::{.remark}
Note: we only needed to check overlaps because of exactness of the following sequence:
\[
0 \to \mcf(U) \to \prod_{i\in I} \mcf(U_i) \to \prod_{i < j} \mcf(U_{ij}) \to \cdots
.\]
:::

## Inverse image and pushforward

:::{.definition title="Pushforward and inverse image sheaves"}
Let $f\in \Top(X, Y)$, let $\mcf \in \Sh(X)$ and define the **pushforward sheaf** $f_* \mcf \in \Sh(Y)$ by
\[
f_*\mcf(V) \da \mcf( f\inv(V))
.\]

The **inverse image** sheaf is define as 
\[
(f\inv \mcf)(U) \da \lim_{\text{open } V\contains f(U)} F(V)
.\]
:::

:::{.remark}
The inverse image sheaf generalizes stalks, recovering $\mcf_p$ when $f(U) = p$.
Note that $f(U)$ need not be open unless $f$ is an open map, and checking that $f(U)$ is (co?)final in the system $\ts{\text{open } V\contains f(U)}$ yields
\[
(f \inv\mcf)(U) = \mcf(f(U))
.\]

:::

:::{.warnings}
We will have a notion of $f^*$, but this will not generally be the pullback!
:::

:::{.exercise title="?"}
Show that $f_* \mcf$ makes sense precisely because $f$ is continuous.
Check that $f_* \mcf$ satisfies the sheaf axioms.
Use that the set of opens of the form $f\inv(V)$ are e.g. closed under intersections, and thus inherit all of the sheaf axioms from $\mcf$.
:::