## 6.22 Continuous maps and abelian sheaves

Let $f : X \to Y$ be a continuous map. We claim there are functors

\begin{eqnarray*} f_* : \textit{PAb}(X) & \longrightarrow & \textit{PAb}(Y) \\ f_* : \textit{Ab}(X) & \longrightarrow & \textit{Ab}(Y) \\ f_ p : \textit{PAb}(Y) & \longrightarrow & \textit{PAb}(X) \\ f^{-1} : \textit{Ab}(Y) & \longrightarrow & \textit{Ab}(X) \end{eqnarray*}

with similar properties to their counterparts in Section 6.21. To see this we argue in the following way.

Each of the functors will be constructed in the same way as the corresponding functor in Section 6.21. This works because all the colimits in that section are directed colimits (but we will work through it below).

First off, given an abelian presheaf $\mathcal{F}$ on $X$ and an abelian presheaf $\mathcal{G}$ on $Y$ we define

\begin{eqnarray*} f_*\mathcal{F}(V) & = & \mathcal{F}(f^{-1}(V)) \\ f_ p\mathcal{G}(U) & = & \mathop{\mathrm{colim}}\nolimits _{f(U) \subset V} \mathcal{G}(V) \end{eqnarray*}

as abelian groups. The restriction mappings are the same as the restriction mappings for presheaves of sets (and they are all homomorphisms of abelian groups).

The assignments $\mathcal{F} \mapsto f_*\mathcal{F}$ and $\mathcal{G} \to f_ p\mathcal{G}$ are functors on the categories of presheaves of abelian groups. This is clear, as (for example) a map of abelian presheaves $\mathcal{G}_1 \to \mathcal{G}_2$ gives rise to a map of directed systems $\{ \mathcal{G}_1(V)\} _{f(U) \subset V} \to \{ \mathcal{G}_2(V)\} _{f(U) \subset V}$ all of whose maps are homomorphisms and hence gives rise to a homomorphism of abelian groups $f_ p\mathcal{G}_1(U) \to f_ p\mathcal{G}_2(U)$.

The functors $f_*$ and $f_ p$ are adjoint on the category of presheaves of abelian groups, i.e., we have

$\mathop{Mor}\nolimits _{\textit{PAb}(X)}(f_ p\mathcal{G}, \mathcal{F}) = \mathop{Mor}\nolimits _{\textit{PAb}(Y)}(\mathcal{G}, f_*\mathcal{F}).$

To prove this, note that the map $i_\mathcal {G} : \mathcal{G} \to f_* f_ p\mathcal{G}$ from the proof of Lemma 6.21.3 is a map of abelian presheaves. Hence if $\psi : f_ p\mathcal{G} \to \mathcal{F}$ is a map of abelian presheaves, then the corresponding map $\mathcal{G} \to f_*\mathcal{F}$ is the map $f_*\psi \circ i_\mathcal {G} : \mathcal{G} \to f_* f_ p \mathcal{G} \to f_* \mathcal{F}$ is also a map of abelian presheaves. For the other direction we point out that the map $c_\mathcal {F} : f_ p f_* \mathcal{F} \to \mathcal{F}$ from the proof of Lemma 6.21.3 is a map of abelian presheaves as well (since it is made out of restriction mappings of $\mathcal{F}$ which are all homomorphisms). Hence given a map of abelian presheaves $\varphi : \mathcal{G} \to f_*\mathcal{F}$ the map $c_\mathcal {F} \circ f_ p\varphi : f_ p\mathcal{G} \to \mathcal{F}$ is a map of abelian presheaves as well. Since these constructions $\psi \mapsto f_*\psi$ and $\varphi \mapsto c_\mathcal {F} \circ f_ p\varphi$ are inverse to each other as constructions on maps of presheaves of sets we see they are also inverse to each other on maps of abelian presheaves.

If $\mathcal{F}$ is an abelian sheaf on $Y$, then $f_*\mathcal{F}$ is an abelian sheaf on $X$. This is true because of the definition of an abelian sheaf and because this is true for sheaves of sets, see Lemma 6.21.1. This defines the functor $f_*$ on the category of abelian sheaves.

We define $f^{-1}\mathcal{G} = (f_ p\mathcal{G})^\#$ as before. Adjointness of $f_*$ and $f^{-1}$ follows formally as in the case of presheaves of sets. Here is the argument:

\begin{eqnarray*} \mathop{Mor}\nolimits _{\textit{Ab}(X)}(f^{-1}\mathcal{G}, \mathcal{F}) & = & \mathop{Mor}\nolimits _{\textit{PAb}(X)}(f_ p\mathcal{G}, \mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{PAb}(Y)}(\mathcal{G}, f_*\mathcal{F}) \\ & = & \mathop{Mor}\nolimits _{\textit{Ab}(Y)}(\mathcal{G}, f_*\mathcal{F}) \end{eqnarray*}

Lemma 6.22.1. Let $f : X \to Y$ be a continuous map.

1. Let $\mathcal{G}$ be an abelian presheaf on $Y$. Let $x \in X$. The bijection $\mathcal{G}_{f(x)} \to (f_ p\mathcal{G})_ x$ of Lemma 6.21.4 is an isomorphism of abelian groups.

2. Let $\mathcal{G}$ be an abelian sheaf on $Y$. Let $x \in X$. The bijection $\mathcal{G}_{f(x)} \to (f^{-1}\mathcal{G})_ x$ of Lemma 6.21.5 is an isomorphism of abelian groups.

Proof. Omitted. $\square$

Given a continuous map $f : X \to Y$ and sheaves of abelian groups $\mathcal{F}$ on $X$, $\mathcal{G}$ on $Y$, the notion of an $f$-map $\mathcal{G} \to \mathcal{F}$ of sheaves of abelian groups makes sense. We can just define it exactly as in Definition 6.21.7 (replacing maps of sets with homomorphisms of abelian groups) or we can simply say that it is the same as a map of abelian sheaves $\mathcal{G} \to f_*\mathcal{F}$. We will use this notion freely in the following. The group of $f$-maps between $\mathcal{G}$ and $\mathcal{F}$ will be in canonical bijection with the groups $\mathop{Mor}\nolimits _{\textit{Ab}(X)}(f^{-1}\mathcal{G}, \mathcal{F})$ and $\mathop{Mor}\nolimits _{\textit{Ab}(Y)}(\mathcal{G}, f_*\mathcal{F})$.

Composition of $f$-maps is defined in exactly the same manner as in the case of $f$-maps of sheaves of sets. In addition, given an $f$-map $\mathcal{G} \to \mathcal{F}$ as above, the induced maps on stalks

$\varphi _ x : \mathcal{G}_{f(x)} \longrightarrow \mathcal{F}_ x$

are abelian group homomorphisms.

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