Topics in Abstract Algebra/Sheaf theory

Sheaf theory

We say $ℱ$ is a pre-sheaf on a topological space $X$ if

• (i) $ℱ(U)$ is an abelian group for every open subset $U\subset X$
• (ii) For each inclusion $U\hookrightarrow V$, we have the group morphism ${\rho }_{V,U}:ℱ(V)\to ℱ(U)$ 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 $f_j\in ℱ(U_j)$ are such that $f_j=f_k$ in $U_j\cap U_k$, then there exists a unique $f\in ℱ(U)$ such that $f{|}_{U_j}=f_j$ for all $j$.

Note that the uniqueness implies that if $f,g\in ℱ(U)$ 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., $𝐑$). Let $ℱ(U)$ be the set of all continuous maps from open subsets $U\subset X$ to $G$. Then $ℱ$ forms a sheaf. In particular, suppose the topology for $G$ is discrete. Then $ℱ$ is called a constant sheaf.

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

${\varphi }_U\circ {\rho }_{V,U}={\rho }_{V,U}\circ {\varphi }_V$

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

Define $(\ker \varphi )(U)=\ker {\varphi }_U$ for each open subset $U$. $\ker \varphi$ is then a sheaf. In fact, suppose $f_j\in \ker {\varphi }_{U_j}$. Then there is $f\in ℱ(U)$ such that $f{|}_{U_j}=f_j$. But since

$({\varphi }_{U}f){|}_{U_j}={\varphi }_{U_j}(f{|}_{U_j})={\varphi }_{U_j}{f}_j=0$

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

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

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

4 Theorem Suppose

$0⟶{ℱ}^0⟶{ℱ}^1⟶{ℱ}^2⟶0$

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

$0⟶{\mathrm{\Gamma }}_Z(U,{ℱ}^0)⟶{\mathrm{\Gamma }}_Z(U,{ℱ}^1)⟶{\mathrm{\Gamma }}_Z(U,{ℱ}^2)$

is exact. Furthermore, ${\mathrm{\Gamma }}_Z(U,{ℱ}^1)\to {\mathrm{\Gamma }}_Z(U,{ℱ}^2)$ is surjective if ${ℱ}^0$ is flabby.
Proof: That the kernel of $\ker {ℱ}^0⟶{ℱ}^1$ is trivial means that $\ker {ℱ}^0(U)⟶{ℱ}^1(U)$ has trivial kernel for any $U$. Thus the first map is clear. Next, denoting ${ℱ}^1\to {ℱ}^2$ by $d$, suppose $f\in {ℱ}^1(U)$ with $df=0$. Then there exists an open cover $U_j$ of $U$ and $u_j\in ℱ(U_j)$ such that $d{u}_j=f{|}_{U_j}$. Since $d{u}_j=f=d{u}_k$ 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 $u\in ℱ(U)$ such that $du=f$. Finally, to show that the last map is surjective, let $f\in {ℱ}^2(U)$, and $\mathrm{\Omega }=\{(U,u)|du=f{|}_U\}$. If $\{(U_j,u_j)|j\in J\}\subset \mathrm{\Omega }$ is totally ordered, then let $U={\cup }_j{U}_j$. Since $u_j$ agree on overlaps by totally ordered-ness, there is $u\in ℱ(U)$ 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 $d{u}_1=f{|}_{U_1}$. Since $d(u_0-u_1)=0$ in $U_0\cap U_1$, by the early part of the proof, there exists $a\in {ℱ}^0(U_0\cap U_1)$ with $da=u_0-u_1$. Then $d(u_1+da)=d{u}_1=f{|}_{U_1}$ (so $(U_1,u_1)\in \mathrm{\Omega }$) 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 $d{u}_0=f$. $◻$

4 Corollary

$0⟶{ℱ}^0⟶{ℱ}^1⟶{ℱ}^2⟶0$

is exact if and only if

$0⟶{ℱ}_p^0⟶{ℱ}_p^1⟶{ℱ}_p^2⟶0$

is exact for every $p\in X$.

Suppose $f:X\to Y$ is a continuous map. The sheaf $f_{\ast }ℱ$ (called the pushforward of $ℱ$ by $f$) is defined by $f_{\ast }ℱ(U)=ℱ(f^{-1}(U))$ for an open subset $U\subset Y$. Suppose $f:Y\to X$ is a continuous map. The sheaf $f^{-1}ℱ$ is then defined by $f^{-1}ℱ(U)=$ the sheafification of the presheaf $U\mapsto \underset{V\supset f(U)}{\underrightarrow{\lim }}ℱ(V)$ 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 $f^{-1}{f}_{\ast }ℱ(U)$ consists of elements $f$ in $ℱ(f^{-1}(V))$ where $V\supset f(U)$. Since $f^{-1}(V)\supset U$, we find a map

$f^{-1}{f}_{\ast }ℱ\to ℱ$

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