18.4 Free abelian presheaves
In order to prepare notation for the following definition, let us agree to denote the free abelian group on a set $S$ as1 $\mathbf{Z}[S] = \bigoplus _{s \in S} \mathbf{Z}$. It is characterized by the property
\[ \mathop{\mathrm{Mor}}\nolimits _{\textit{Ab}}(\mathbf{Z}[S], A) = \mathop{\mathrm{Mor}}\nolimits _{\textit{Sets}}(S, A) \]
In other words the construction $S \mapsto \mathbf{Z}[S]$ is a left adjoint to the forgetful functor $\textit{Ab} \to \textit{Sets}$.
Definition 18.4.1. Let $\mathcal{C}$ be a category. Let $\mathcal{G}$ be a presheaf of sets. The free abelian presheaf $\mathbf{Z}_\mathcal {G}$ on $\mathcal{G}$ is the abelian presheaf defined by the rule
\[ U \longmapsto \mathbf{Z}[\mathcal{G}(U)]. \]
In the special case $\mathcal{G} = h_ X$ of a representable presheaf associated to an object $X$ of $\mathcal{C}$ we use the notation $\mathbf{Z}_ X = \mathbf{Z}_{h_ X}$. In other words
\[ \mathbf{Z}_ X(U) = \mathbf{Z}[\mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U, X)]. \]
This construction is clearly functorial in the presheaf $\mathcal{G}$. In fact it is adjoint to the forgetful functor $\textit{PAb}(\mathcal{C}) \to \textit{PSh}(\mathcal{C})$. Here is the precise statement.
Lemma 18.4.2. Let $\mathcal{C}$ be a category. Let $\mathcal{G}$, $\mathcal{F}$ be a presheaves of sets. Let $\mathcal{A}$ be an abelian presheaf. Let $U$ be an object of $\mathcal{C}$. Then we have
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{F}) & = \mathcal{F}(U), \\ \mathop{\mathrm{Mor}}\nolimits _{\textit{PAb}(\mathcal{C})}(\mathbf{Z}_\mathcal {G}, \mathcal{A}) & = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(\mathcal{G}, \mathcal{A}), \\ \mathop{\mathrm{Mor}}\nolimits _{\textit{PAb}(\mathcal{C})}(\mathbf{Z}_ U, \mathcal{A}) & = \mathcal{A}(U). \end{align*}
All of these equalities are functorial.
Proof.
Omitted.
$\square$
Lemma 18.4.3. Let $\mathcal{C}$ be a category. Let $I$ be a set. For each $i \in I$ let $\mathcal{G}_ i$ be a presheaf of sets. Then
\[ \mathbf{Z}_{\coprod _ i \mathcal{G}_ i} = \bigoplus \nolimits _{i \in I} \mathbf{Z}_{\mathcal{G}_ i} \]
in $\textit{PAb}(\mathcal{C})$.
Proof.
Omitted.
$\square$
Comments (0)