# Topics in Abstract Algebra/Sheaf theory

## Sheaf theory

We say ${\displaystyle {\mathcal {F}}}$ is a pre-sheaf on a topological space ${\displaystyle X}$ if

• (i) ${\displaystyle {\mathcal {F}}(U)}$ is an abelian group for every open subset ${\displaystyle U\subset X}$
• (ii) For each inclusion ${\displaystyle U\hookrightarrow V}$, we have the group morphism ${\displaystyle \rho _{V,U}:{\mathcal {F}}(V)\to {\mathcal {F}}(U)}$ such that
${\displaystyle \rho _{U,U}}$ is the identity and ${\displaystyle \rho _{W,U}=\rho _{V,U}\circ \rho _{W,V}}$ for any inclusion ${\displaystyle U\hookrightarrow V\hookrightarrow W}$

A pre-sheaf is called a sheaf if the following "gluing axiom" holds:

For each open subset ${\displaystyle U}$ and its open cover ${\displaystyle U_{j}}$, if ${\displaystyle f_{j}\in {\mathcal {F}}(U_{j})}$ are such that ${\displaystyle f_{j}=f_{k}}$ in ${\displaystyle U_{j}\cap U_{k}}$, then there exists a unique ${\displaystyle f\in {\mathcal {F}}(U)}$ such that ${\displaystyle f|_{U_{j}}=f_{j}}$ for all ${\displaystyle j}$.

Note that the uniqueness implies that if ${\displaystyle f,g\in {\mathcal {F}}(U)}$ and ${\displaystyle f|_{U_{j}}=g|_{U_{j}}}$ for all ${\displaystyle j}$, then ${\displaystyle f=g}$. In particular, ${\displaystyle f|_{U_{j}}=0}$ for all ${\displaystyle j}$ implies ${\displaystyle f=0}$.

4 Example: Let ${\displaystyle G}$ be a topological group (e.g., ${\displaystyle \mathbf {R} }$). Let ${\displaystyle {\mathcal {F}}(U)}$ be the set of all continuous maps from open subsets ${\displaystyle U\subset X}$ to ${\displaystyle G}$. Then ${\displaystyle {\mathcal {F}}}$ forms a sheaf. In particular, suppose the topology for ${\displaystyle G}$ is discrete. Then ${\displaystyle {\mathcal {F}}}$ is called a constant sheaf.

Given sheaves ${\displaystyle {\mathcal {F}}}$ and ${\displaystyle {\mathcal {G}}}$, a sheaf morphism ${\displaystyle \phi :{\mathcal {F}}\to {\mathcal {G}}}$ is a collection of group morphisms ${\displaystyle \phi _{U}:{\mathcal {F}}(U)\to {\mathcal {G}}(U)}$ satisfying: for every open subset ${\displaystyle U\subset V}$,

${\displaystyle \phi _{U}\circ \rho _{V,U}=\rho _{V,U}\circ \phi _{V}}$

where the first ${\displaystyle \rho _{V,U}}$ is one that comes with ${\displaystyle {\mathcal {F}}}$ and the second ${\displaystyle {\mathcal {G}}}$.

Define ${\displaystyle (\operatorname {ker} \phi )(U)=\operatorname {ker} \phi _{U}}$ for each open subset ${\displaystyle U}$. ${\displaystyle \operatorname {ker} \phi }$ is then a sheaf. In fact, suppose ${\displaystyle f_{j}\in \operatorname {ker} \phi _{U_{j}}}$. Then there is ${\displaystyle f\in {\mathcal {F}}(U)}$ such that ${\displaystyle f|_{U_{j}}=f_{j}}$. But since

${\displaystyle (\phi _{U}f)|_{U_{j}}=\phi _{U_{j}}(f|_{U_{j}})=\phi _{U_{j}}f_{j}=0}$

for all ${\displaystyle j}$, we have ${\displaystyle \phi _{U}f=0}$. Unfortunately, ${\displaystyle \operatorname {im} \phi }$ does not turn out to be a sheaf if it is defined in the same way. We thus define ${\displaystyle (\operatorname {im} \phi )(U)}$ to be the set of all ${\displaystyle f\in {\mathcal {G}}(U)}$ such that there is an open cover ${\displaystyle U_{j}}$ of ${\displaystyle U}$ such that ${\displaystyle f|_{U_{j}}}$ is in the image of ${\displaystyle \phi _{U_{j}}}$. This is a sheaf. In fact, as before, let ${\displaystyle f\in {\mathcal {G}}(U)}$ be such that ${\displaystyle f|_{U_{j}}\in \operatorname {im} \phi _{U_{j}}}$. Then we have an open cover of ${\displaystyle U}$ such that ${\displaystyle f}$ restricted to each member ${\displaystyle V}$ of the cover is in the image of ${\displaystyle \phi _{V}}$.

Let ${\displaystyle {\mathcal {F}}^{0},{\mathcal {F}}^{1},{\mathcal {F}}^{2}}$ be sheaves on the same topological space.

A sheaf ${\displaystyle {\mathcal {F}}}$ on ${\displaystyle X}$ is said to be flabby if ${\displaystyle \rho _{X,U}:{\mathcal {F}}(X)\to {\mathcal {F}}(U)}$ is surjective. Let ${\displaystyle {\mathcal {F}}_{p}=\lim _{U\ni p}{\mathcal {F}}(U)}$, and, for each ${\displaystyle f\in {\mathcal {F}}(U)}$, define ${\displaystyle \operatorname {supp} f=\{x\in U|f|_{p}\neq 0\}}$. ${\displaystyle \operatorname {supp} f}$ is closed since ${\displaystyle f|_{p}=0}$ implies ${\displaystyle p}$ has a neighborhood of ${\displaystyle U}$ such that ${\displaystyle f|_{q}=0}$ for every ${\displaystyle q\in U}$. Define ${\displaystyle \operatorname {Supp} {\mathcal {F}}=\{x\in X|{\mathcal {F}}_{x}\neq 0\}}$. In particular, if ${\displaystyle i:Z\hookrightarrow X}$ is a closed subset and ${\displaystyle \operatorname {Supp} {\mathcal {F}}\subset Z}$, then the natural map ${\displaystyle {\mathcal {F}}\to i_{*}i^{-1}{\mathcal {F}}}$ is an isomorphism.

4 Theorem Suppose

${\displaystyle 0\longrightarrow {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}\longrightarrow {\mathcal {F}}^{2}\longrightarrow 0}$

is exact. Then, for every open subset ${\displaystyle U}$

${\displaystyle 0\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{0})\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{1})\longrightarrow \Gamma _{Z}(U,{\mathcal {F}}^{2})}$

is exact. Furthermore, ${\displaystyle \Gamma _{Z}(U,{\mathcal {F}}^{1})\to \Gamma _{Z}(U,{\mathcal {F}}^{2})}$ is surjective if ${\displaystyle {\mathcal {F}}^{0}}$ is flabby.
Proof: That the kernel of ${\displaystyle \operatorname {ker} {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}}$ is trivial means that ${\displaystyle \operatorname {ker} {\mathcal {F}}^{0}(U)\longrightarrow {\mathcal {F}}^{1}(U)}$ has trivial kernel for any ${\displaystyle U}$. Thus the first map is clear. Next, denoting ${\displaystyle {\mathcal {F}}^{1}\to {\mathcal {F}}^{2}}$ by ${\displaystyle d}$, suppose ${\displaystyle f\in {\mathcal {F}}^{1}(U)}$ with ${\displaystyle df=0}$. Then there exists an open cover ${\displaystyle U_{j}}$ of ${\displaystyle U}$ and ${\displaystyle u_{j}\in {\mathcal {F}}(U_{j})}$ such that ${\displaystyle du_{j}=f|_{U_{j}}}$. Since ${\displaystyle du_{j}=f=du_{k}}$ in ${\displaystyle U_{j}\cap U_{k}}$ and ${\displaystyle d_{U_{j}\cap U_{k}}}$ is injective by the early part of the proof, we have ${\displaystyle u_{j}=u_{k}}$ in ${\displaystyle U_{j}\cap U_{k}}$ and so we get ${\displaystyle u\in {\mathcal {F}}(U)}$ such that ${\displaystyle du=f}$. Finally, to show that the last map is surjective, let ${\displaystyle f\in {\mathcal {F}}^{2}(U)}$, and ${\displaystyle \Omega =\{(U,u)|du=f|_{U}\}}$. If ${\displaystyle \{(U_{j},u_{j})|j\in J\}\subset \Omega }$ is totally ordered, then let ${\displaystyle U=\cup _{j}U_{j}}$. Since ${\displaystyle u_{j}}$ agree on overlaps by totally ordered-ness, there is ${\displaystyle u\in {\mathcal {F}}(U)}$ with ${\displaystyle u|_{U_{j}}=u_{j}}$. Thus, ${\displaystyle (U,u)}$ is an upper bound of the collection ${\displaystyle (U_{j},u_{j})}$. By Zorn's Lemma, we then find a maximal element ${\displaystyle (U_{0},u_{0})}$. We claim ${\displaystyle U_{0}=U}$. Suppose not. Then there exists ${\displaystyle (U_{1},u_{1})}$ with ${\displaystyle du_{1}=f|_{U_{1}}}$. Since ${\displaystyle d(u_{0}-u_{1})=0}$ in ${\displaystyle U_{0}\cap U_{1}}$, by the early part of the proof, there exists ${\displaystyle a\in {\mathcal {F}}^{0}(U_{0}\cap U_{1})}$ with ${\displaystyle da=u_{0}-u_{1}}$. Then ${\displaystyle d(u_{1}+da)=du_{1}=f|_{U_{1}}}$ (so ${\displaystyle (U_{1},u_{1})\in \Omega }$) while ${\displaystyle u_{1}+da=u_{0}}$ in ${\displaystyle U_{0}\cap U_{1}}$. This contradicts the maximality of ${\displaystyle (U_{0},u_{0})}$. Hence, we conclude ${\displaystyle U_{0}=U}$ and so ${\displaystyle du_{0}=f}$. ${\displaystyle \square }$

4 Corollary

${\displaystyle 0\longrightarrow {\mathcal {F}}^{0}\longrightarrow {\mathcal {F}}^{1}\longrightarrow {\mathcal {F}}^{2}\longrightarrow 0}$

is exact if and only if

${\displaystyle 0\longrightarrow {\mathcal {F}}_{p}^{0}\longrightarrow {\mathcal {F}}_{p}^{1}\longrightarrow {\mathcal {F}}_{p}^{2}\longrightarrow 0}$

is exact for every ${\displaystyle p\in X}$.

Suppose ${\displaystyle f:X\to Y}$ is a continuous map. The sheaf ${\displaystyle f_{*}{\mathcal {F}}}$ (called the pushforward of ${\displaystyle {\mathcal {F}}}$ by ${\displaystyle f}$) is defined by ${\displaystyle f_{*}{\mathcal {F}}(U)={\mathcal {F}}(f^{-1}(U))}$ for an open subset ${\displaystyle U\subset Y}$. Suppose ${\displaystyle f:Y\to X}$ is a continuous map. The sheaf ${\displaystyle f^{-1}{\mathcal {F}}}$ is then defined by ${\displaystyle f^{-1}{\mathcal {F}}(U)=}$ the sheafification of the presheaf ${\displaystyle U\mapsto \varinjlim _{V\supset f(U)}{\mathcal {F}}(V)}$ where ${\displaystyle V}$ is an open subset of ${\displaystyle X}$. The two are related in the following way. Let ${\displaystyle U\subset X}$ be an open subset. Then ${\displaystyle f^{-1}f_{*}{\mathcal {F}}(U)}$ consists of elements ${\displaystyle f}$ in ${\displaystyle {\mathcal {F}}(f^{-1}(V))}$ where ${\displaystyle V\supset f(U)}$. Since ${\displaystyle f^{-1}(V)\supset U}$, we find a map

${\displaystyle f^{-1}f_{*}{\mathcal {F}}\to {\mathcal {F}}}$

by sending ${\displaystyle f}$ to ${\displaystyle f|_{U}}$. The map is well-defined for it doesn't depend on the choice of ${\displaystyle V}$.