sheaf 1 - Time Ordering

The main reference: Hartshorne.

A topological space $X$ can be seen as a category if we take open sets as its objects and inclusion maps as its morphisms, i.e. $\operatorname{Hom}_{X}(U,V)=\begin{cases} \bigl\{i_{UV}:U\hookrightarrow V\bigr\},& \text{if }U\subset V\subset X\text{;}\\ \varnothing,&\text{otherwise}. \end{cases}$

Definition 1: A presheaf is a functor $\mathcal{F}:X^\circ\to \textit{AG}$ , where $X^\circ$ is the dual category of a topological space $X$ and $\textit{AG}$ is the category of Abelian groups. In other words, $\mathcal{F}:X\to \textit{AG}$ is a contravariant functor. The elements $s\in \mathcal{F}(U)$ are called the sections on $U$ . Denote that $\rho_{UV}:=\mathcal{F}(i_{UV}):\mathcal{F}(V)\to\mathcal{F}(U)$ and $\rho_{UV}(s)=s|_V$ for $s \in \mathcal{F}(U)$ .

The morphisms $\varphi$ between two presheaf $\mathcal{F}$ and $\mathcal{G}$ is just the morphisms between two functors, that is, there exist a family of morphisms $\varphi(U):\mathcal{F}(U)\to \mathcal{G}(U)$ such that the following diagram is commutative.

Definition 2: If $\mathcal{F}$ is a presheaf on $X$ , and if $p$ is a point on $X$ , we define the stalk $\mathcal{F}_p\in \textit{AG}$ of $\mathcal{F}$ at $p$ to be the direct limit of the group $\mathcal{F}(U)$ for all $U \subset X$ containing $p$ , via the restriction maps $\rho$ . We can directly construct $\mathcal{F}_p=\underrightarrow{\lim}_{U\ni p}\mathcal{F}(U)$ that $\mathcal{F}_p=\bigl\{\langle U,s\rangle: p\in U\subset X, s\in \mathcal{F}(U)\bigr\}/\sim,$ where $\sim$ is defined as follows: Suppose $s\in\mathcal{F}(U)$ and $t\in\mathcal{F}(V)$ , if there exists a $W\subset U\cap V\neq \varnothing$ such that $s|_W=t|_W$ , then $\langle U,s\rangle=\langle V,t\rangle$ or $s\sim t$ . $\mathcal{F}_p$ is indeed a group since the addition can be defined by $\langle U,s\rangle+\langle V,t\rangle=\langle W,s|_W+t|_W\rangle$ .

Here $\mathcal{F}(U)\ni s\mapsto \langle U,s\rangle=s_p$ defines a family of morphisms $\rho_{Up}:\mathcal{F}(U)\to \mathcal{F}_p$ for any $U\subset X$ . We can then vertify the universal property of direct product that

Now, suppose $\varphi:\mathcal{F} \to \mathcal{G}$ is a functor. The diagram

is commutative, then the universal property of direct product

gives the existence of the morphism $\varphi_p:\mathcal{F}_p\to \mathcal{G}_p$ , and the diagram

is commutative.

Definition 3: A sheaf is a presheaf that for any open set $U\subset X$ , the complex $0\rightarrow \mathcal{F}(U) \xrightarrow{d_0}\prod_{i\in I}\mathcal{F}(U_i) \xrightarrow{d_1}\prod_{i,j\in I}\mathcal{F}(U_i\cap U_j)$ is exact for any open cover $\{U_i\}$ of $U$ , where $\begin{array}{cccl} d_0:&s&\mapsto& \displaystyle{\prod_{i\in I}s|_{U_i}},\\ d_1:&\displaystyle{\prod_{i\in I}s_i}&\mapsto& \displaystyle{\prod_{i,j\in I}\bigl(s_i|_{U_i\cap U_j}-s_j|_{U_i\cap U_j}\bigr)}. \end{array}$ The definition can be rewritten as: For any open cover $\{U_i\}$ of any open set $U\subset X$ ,

If $\forall i\in I$ , $s|_{U_i}=0$ , then $s=0$ .
If $\forall i,j\in I$ , $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ , then there’s a section $s\in\mathcal{F}(U)$ such that $s|_{U_i}=s_i$ .

It is not so difficult to vertify that this definition is equivalent to the old one.

The following proposition (which would be false for presheves) illustrates the local nature of a sheaf.

Propostion 1: Suppose $\varphi:\mathcal{F}\to\mathcal{G}$ is a morphism of sheaves on a topological space $X$ . Then $\varphi$ is a isomorphism if and only if the induced map on the stalk $\varphi_p:\mathcal{F}_p\to \mathcal{G}_p$ is an isomorphism for every $p\in X$ .

Proof. p.63 on Hartshorne.

Definition 4: Let $\varphi:\mathcal{F}\to\mathcal{G}$ be a morphism of presheaves. We define the presheaf kernel of $\varphi$ , presheaf of cokernel of $\varphi$ , and presheaf image of $\varphi$ to be the presheaves given by $U\mapsto \ker(\varphi(U))$ , $U\to \operatorname{coker}(\varphi(U))$ , and $U\mapsto \operatorname{Im}(\varphi(U))$ respectively. Then Proposition 2: If $\varphi:\mathcal{F}\to\mathcal{G}$ is a morphism of sheaves, then the presheaf $U\mapsto \ker(\varphi(U))$ is a sheaf.

Proof. Let

\{U_i\}

be an open cover of

U

, and

s_i

is local section on

U_i

Suppose $s\in \ker(\varphi(U))$ and $s|_{U_i}=0$ , since $\mathcal{F}$ is a sheaf, $s=0$ .

Suppose $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ , we need to show that there exists a global section $s\in \ker(\varphi(U))$ such that $s|_{U_i}=s_i$ . Since $\mathcal{F}$ is a sheaf, it’s nature that there’s $s\in \mathcal{F}(U)$ such that $s|_{U_i}=s_i$ . The last thing to vetify is $\varphi(U)(s)=0$ . Restrict $\varphi(U)(s)$ on $U_i$ , then $\rho'_{UU_i}\circ \varphi(U)(s)=\varphi(U_i)(\rho_{UU_i}s)=\varphi(U_i)(s_i)=0,$ so $\varphi(U)(s)\in \mathcal{G}(U)$ vanishes locally. Since $\mathcal{G}$ is a sheaf, it also vanishes globally, i.e. $\varphi(U)(s)=0$ .

Thus

U\mapsto \ker(\varphi(U))

is a sheaf.

However, the presheaves $\operatorname{coker}(\varphi)$ and $\operatorname{Im}(\varphi)$ need not to be sheaves. Actually, the key point in the proof above is that $\ker$ is compatible with the sheaf property of $\mathcal{G}$ . Then we come to an important notion of a sheaf associated to a presheaf, i.e. sheafification.

Roughly speaking, the sheafification of a presheaf $\mathcal{F}$ is the “smallest” sheaf with the same stalks as $\mathcal{F}$ . Because of the “smallest”, sheafification should have the universal property.

Propostion 3: Given a presheaf $\mathcal{F}$ , there is a sheaf $\mathcal{F}^+$ and a morphism $\theta$ make the diagram
commutative for any sheaf $\mathcal{G}$ . $\mathcal{F}^+$ is called the sheaf associated to the preshead $\mathcal{F}$ or the sheafification of $\mathcal{F}$ .

Proof. We construct the sheaf

\mathcal{F}^+

as follows. For any open set

U

, let

\mathcal{F}^+(U)

be the set of functions

s:U\to \cup_{p\in U}\mathcal{F}_p

, such that

for each $p\in U$ , $s(p)\in \mathcal{F}_p$ , and

for each $p\in U$ , there is a neighborhood $V\subset U$ of $P$ , and an element $t\in\mathcal{F}(V)$ , such that $\forall q\in V$ , $s(q)=t_q:=t^+(q)$ .

The addition on $\mathcal{F}^+(U)$ is that $(s+t)(p)=s(p)+t(p)$ , so $\mathcal{F}^+(U)$ is indeed a group. If $V\subset U$ , there’s a nature map (function restriction) $i_{UV}:\mathcal{F}^+(U)\to \mathcal{F}^+(V)$ such that $i_{UV}(s)=s|_V$ , so $\mathcal{F}^+$ is a presheaf. Let $\{U_i\}$ be a open cover of $U$ and $s_i\in \mathcal{F}^+(U_i)$ be local sections. If $s_i|_{U_i\cap U_j}=s_j|_{U_i\cap U_j}$ , we can define a function $s:U\to \cup_{p\in U}\mathcal{F}_p$ by setting $s|_{U_i}=s_i$ . If $s|_{U_i}=0$ for all $i\in I$ , then $s=0$ since it is a function. Thus $\mathcal{F}^+$ is a sheaf.

For each $s\in \mathcal{F}(U)$ , we can associate it a section $s^+\in \mathcal{F}^+(U)$ by $s^+(p)=s_p$ , then there’s a morphism $\theta(U):s\mapsto s^+$ . $\forall s\in \mathcal{F}(U)$ , since $i_{UV}\bigl(\theta(U)(s)\bigr)=i_{UV}(s^+)=s^+|_{V}$ and $\theta(V)\bigl(\rho_{UV}(s)\bigr)=\theta(V)(s|_V)=s^+|_{V}.$ $\theta$ is a morphism.

Let

\bar{s}\in \mathcal{F}^+(U)

, because of the construction of

\mathcal{F}^+

, we can find an open cover

\{U_i\}

U

such that

\bar{s}|_{U_i}=\bar{s}^+_i

, where

\bar{s}_i\in \mathcal{F}(U_i)

. Firstly define

\psi(U_i):\bar{s}|_{U_i}\mapsto \varphi(U_i)(\bar{s}_i)

, and we can use the sheaf condition of

\mathcal{G}

to get a global section

s'

U

such that

s'|_{U_i}=\varphi(U_i)(\bar{s}_i)

. Then, define

\psi(U):\bar{s}\mapsto s'

, and

\psi

will become the morphism

\psi:\mathcal{F}^+\to \mathcal{G}

. Finally, because of the construction of

\psi

, for any

s^+\in \mathcal{F}^+(U)

\psi(U)(s^+)=\varphi(U)(s)

, so that

\psi(U)(\theta(s))=\psi(U)(s^+)=\varphi(U)(s)

makes the diagram commutative.

Proposition 4: $\mathcal{F}_p\cong \mathcal{F}^+_p$ , so if $\mathcal{F}$ is a sheaf, then $\mathcal{F}\cong \mathcal{F}^+$ .

Proof. There's a morphism

\theta_p:\mathcal{F}_p\to \mathcal{F}^+_p

, such that

\theta_p(\langle U,s\rangle)=\langle U,s^+\rangle

It is injective. If $\theta_p(\langle U,s\rangle)=\langle V,0\rangle$ , then $s^+|_V=0$ , and $s_p=s^+(p)=0$ .

It is surjective.

\forall \langle U,\bar{s}\rangle\in \mathcal{F}^+_p

, there exists an open subset

V

and

t\in \mathcal{F}(V)

such that

\langle U,\bar{s}\rangle=\langle V,t^+\rangle

, then

\theta_p(\langle V,t\rangle)=\langle U,\bar{s}\rangle

Definiton 5: If $\varphi:\mathcal{F}\to\mathcal{G}$ is a morphism of sheaves, we define the kernel(repsectively cokernel, image) of $\varphi$ , denoted $\ker \varphi$ (repsectively $\operatorname{coker} \varphi$ , $\operatorname{Im} \varphi$ ), to be the sheaf associated to the presheaf of kernel(respectively, coker, image) of $\varphi$ .

Definiton 6: We say that a morphism of sheaves $\varphi:\mathcal{F}\to\mathcal{G}$ is injective(respectively, surjective) if $\ker\varphi=0$ (respectively, $\operatorname{Im}\varphi\cong\mathcal{G}$ ).

and

Propositon 5: For any morphism of sheaves $\varphi:\mathcal{F}\to\mathcal{G}$ , $(\ker \varphi)_p=\ker(\varphi_p)$ and $(\operatorname{Im} \varphi)_p=\operatorname{Im}(\varphi_p)$ for each $p$ .

Proof. We will prove it in the following steps:

$\ker(\varphi_p)\subset (\ker \varphi)_p$ Suppose $s_p=\langle U,s\rangle\in \ker(\varphi_p)$ , then $\varphi_p\langle U,s\rangle=\langle U,\varphi(U)s\rangle=0$ , so there exists $W\subset U$ such that $(\varphi(U)s)|_W=\varphi(W)(s|_W)=0,$ thus $s|_W\in \ker(\varphi(W))$ and $s_p=\langle W,s|_W\rangle\in (\ker \varphi)_p$ .
$(\ker \varphi)_p\subset\ker(\varphi_p)$ Suppose $\langle U,s\rangle\in (\ker \varphi)_p$ , then $\varphi_p\langle U,s\rangle=\langle U,\varphi(U)s\rangle=\langle U,0\rangle=0\in \mathcal{G}_p$ , thus $\langle U,s\rangle\in \ker(\varphi_p)$ .
$\operatorname{Im}(\varphi_p)\subset (\operatorname{Im} \varphi)_p$ Suppose $t=\varphi_ps_p\in\operatorname{Im} (\varphi_p)$ and $s_p=\langle U,s\rangle$ , then $t=\varphi_ps_p=\varphi_p\langle U,s\rangle=\langle U,\varphi(U)s\rangle\in (\operatorname{Im} \varphi)_p$ .
$(\operatorname{Im} \varphi)_p\subset \operatorname{Im}(\varphi_p)$ Suppose $t=\langle U,\varphi(U)s\rangle\in (\operatorname{Im} \varphi)_p$ , then $t=\rho'_{Up}\circ\varphi(U)s=\varphi_p s_p \in \operatorname{Im}(\varphi_p)$ .

That's all.

Corollary 1: For any morphism of sheaves $\varphi:\mathcal{F}\to\mathcal{G}$ , it is injective(respectively, surjective) if and only if $\varphi_p$ is injective(respectively, surjective) for all $p$ .

Proof. According to Proposition 1 and Proposition 5,

\ker \varphi=0

if and only if

(\ker \varphi)_p=\ker(\varphi_p)=0

. Similarly,

\operatorname{Im} \varphi\cong \mathcal{G}

if and only if

(\operatorname{Im} \varphi)_p=\operatorname{Im}(\varphi_p)\cong \mathcal{G}_g

Corollary 2: A morphism of sheaves is an isomorphism if and only if it is injective and surjective.

Proof. A morphism of sheaves

\varphi

is an isomorphism if and only if

\varphi_p

is an isomorphism for all

p

. As a morphism of groups,

\varphi_p

is an isomorphism for all

p

if and only if it is injective and surjective for all

p

, and according to Corollay 1, if and only if

\varphi

is injective and surjective.

Definition 7: Let $f:X\to Y$ be a continuous map of topological spoaces. For any sheaf $\mathcal{F}$ on $X$ , we define the direct image sheaf $f_*\mathcal{F}$ on $Y$ by $(f_*\mathcal{F})(U)=\mathcal{F}(f^{-1}(U))$ for any open set $U\subset Y$ . For any sheaf $\mathcal{G}$ on $Y$ , we define the inverse image sheaf $f^{-1}\mathcal{G}$ on $X$ to be the sheaf associated to the presheaf $U\mapsto \underrightarrow{\lim}_{V \supset f(U)}\mathcal{G}(V)$ , where $U$ is any open set in $X$ , and the limit is taken over all open sets $V\subset Y$ containing $f(U)$ .

Buwai Lee

交换图都不会画的魔法师