# Topics in Abstract Algebra/Sheaf theory

## Sheaf theory

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4 Theorem Suppose

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

is exact. Then, for every open subset $U$

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

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

4 Corollary

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

is exact if and only if

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

is exact for every $p\in X$ .

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

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

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