
## 6.16 Exactness and points

In any category we have the notion of epimorphism, monomorphism, isomorphism, etc.

Lemma 6.16.1. Let $X$ be a topological space. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a morphism of sheaves of sets on $X$.

1. The map $\varphi$ is a monomorphism in the category of sheaves if and only if for all $x \in X$ the map $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$ is injective.

2. The map $\varphi$ is an epimorphism in the category of sheaves if and only if for all $x \in X$ the map $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$ is surjective.

3. The map $\varphi$ is an isomorphism in the category of sheaves if and only if for all $x \in X$ the map $\varphi _ x : \mathcal{F}_ x \to \mathcal{G}_ x$ is bijective.

Proof. Omitted. $\square$

It follows that in the category of sheaves of sets the notions epimorphism and monomorphism can be described as follows.

Definition 6.16.2. Let $X$ be a topological space.

1. A presheaf $\mathcal{F}$ is called a subpresheaf of a presheaf $\mathcal{G}$ if $\mathcal{F}(U) \subset \mathcal{G}(U)$ for all open $U \subset X$ such that the restriction maps of $\mathcal{G}$ induce the restriction maps of $\mathcal{F}$. If $\mathcal{F}$ and $\mathcal{G}$ are sheaves, then $\mathcal{F}$ is called a subsheaf of $\mathcal{G}$. We sometimes indicate this by the notation $\mathcal{F} \subset \mathcal{G}$.

2. A morphism of presheaves of sets $\varphi : \mathcal{F} \to \mathcal{G}$ on $X$ is called injective if and only if $\mathcal{F}(U) \to \mathcal{G}(U)$ is injective for all $U$ open in $X$.

3. A morphism of presheaves of sets $\varphi : \mathcal{F} \to \mathcal{G}$ on $X$ is called surjective if and only if $\mathcal{F}(U) \to \mathcal{G}(U)$ is surjective for all $U$ open in $X$.

4. A morphism of sheaves of sets $\varphi : \mathcal{F} \to \mathcal{G}$ on $X$ is called injective if and only if $\mathcal{F}(U) \to \mathcal{G}(U)$ is injective for all $U$ open in $X$.

5. A morphism of sheaves of sets $\varphi : \mathcal{F} \to \mathcal{G}$ on $X$ is called surjective if and only if for every open $U$ of $X$ and every section $s$ of $\mathcal{G}(U)$ there exists an open covering $U = \bigcup U_ i$ such that $s|_{U_ i}$ is in the image of $\mathcal{F}(U_ i) \to \mathcal{G}(U_ i)$ for all $i$.

Lemma 6.16.3. Let $X$ be a topological space.

1. Epimorphisms (resp. monomorphisms) in the category of presheaves are exactly the surjective (resp. injective) maps of presheaves.

2. Epimorphisms (resp. monomorphisms) in the category of sheaves are exactly the surjective (resp. injective) maps of sheaves, and are exactly those maps with are surjective (resp. injective) on all the stalks.

3. The sheafification of a surjective (resp. injective) morphism of presheaves of sets is surjective (resp. injective).

Proof. Omitted. $\square$

Lemma 6.16.4. let $X$ be a topological space. Let $(\mathcal{C}, F)$ be a type of algebraic structure. Suppose that $\mathcal{F}$, $\mathcal{G}$ are sheaves on $X$ with values in $\mathcal{C}$. Let $\varphi : \mathcal{F} \to \mathcal{G}$ be a map of the underlying sheaves of sets. If for all points $x \in X$ the map $\mathcal{F}_ x \to \mathcal{G}_ x$ is a morphism of algebraic structures, then $\varphi$ is a morphism of sheaves of algebraic structures.

Proof. Let $U$ be an open subset of $X$. Consider the diagram of (underlying) sets

$\xymatrix{ \mathcal{F}(U) \ar[r] \ar[d] & \prod _{x \in U} \mathcal{F}_ x \ar[d] \\ \mathcal{G}(U) \ar[r] & \prod _{x \in U} \mathcal{G}_ x }$

By assumption, and previous results, all but the left vertical arrow are morphisms of algebraic structures. In addition the bottom horizontal arrow is injective, see Lemma 6.11.1. Hence we conclude by Lemma 6.15.4, see also Example 6.15.5 $\square$

Short exact sequences of abelian sheaves, etc will be discussed in the chapter on sheaves of modules. See Modules, Section 17.3.

Comment #3011 by David Taylor on

Typo:

In the statement of Definition 6.16.2 (5), on the last line, it should say "...is in the image of $\mathcal{F}(U_i)\rightarrow\mathcal{G}(U_i)$ for all $i$." rather than "...is in the image of $\mathcal{F}(U_i)\rightarrow\mathcal{G}(U)$ for all $i$."

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