# Topics in Abstract Algebra/Sheaf theory

## Sheaf theoryEdit

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 ${\displaystyle V}$ .