How to construct a group-valued sheaf from set-valued sheaf

Question

Suppose that I have a set-valued sheaf $S$ on a site $(\mathcal C, J)$.

Question 1.) Is there a canonical way to turn $S$ into a sheaf with Abelian group values?

I considered the following: for each object $U$ of $\mathcal C$, one can take the free $\mathbb Z$-module, generated by the set $S(u)$, and define the restriction maps on the generators via the restriction maps of $S$. Generally, this will only result in a presheaf, so one applies a sheaffication process.

Question 2.) What would be a sufficient condition on $S$ to ensure that this presheaf is actually a sheaf? I was not able to come up with one myself.

My motivation: I would like to compute the sheaf cohomology of a set-valued sheaf.

Answer 1

Fix a site. Let $\textbf{Psh}$, $\textbf{Sh}$, $\textbf{AbPsh}$, and $\textbf{AbSh}$ be the categories of presheaves/sheaves of sets/abelian groups. We have a commutative diagram of forgetful functors: $$\require{AMScd} \begin{CD} \textbf{AbSh} @>>> \textbf{Sh} \\ @VVV @VVV \\ \textbf{AbPsh} @>>> \textbf{Psh} \end{CD}$$ All of these functors have left adjoints, so if there is anything that deserves to be called the "canonical" sheaf of abelian groups associated with a sheaf of sets, it should be the left adjoint functor $\textbf{Sh} \to \textbf{AbSh}$. It can be computed as you say:

Proposition. The left adjoint functor $\textbf{Sh} \to \textbf{AbSh}$ is isomorphic to the composite $\textbf{Sh} \to \textbf{Psh} \to \textbf{AbPsh} \to \textbf{AbSh}$.

Proof. Let $S$ be a sheaf of sets and let $A$ be a sheaf of abelian groups. Then: $$\begin{align} \textbf{Sh} (S, A) & \cong \textbf{Psh} (S, A) \\ & \cong \textbf{AbPsh} (\mathbb{Z} S, A) \\ & \cong \textbf{AbSh} ((\mathbb{Z} S)^{\sim}, A) \end{align}$$ Here, we have used the fact that $\textbf{Sh} \hookrightarrow \textbf{Psh}$ is fully faithful, $\mathbb{Z} S$ is the result of applying the left adjoint functor $\textbf{Psh} \to \textbf{AbPsh}$ to $S$ (considered as a presheaf), and $(\mathbb{Z} S)^{\sim}$ is the result of applying the left adjoint functor $\textbf{AbPsh} \to \textbf{AbSh}$. 　◼

For completeness, let me note that left adjoint functor $\textbf{Psh} \to \textbf{AbPsh}$ is indeed computed by applying the free abelian group functor componentwise, and that the left adjoint functor $\textbf{AbPsh} \to \textbf{AbSh}$ is "the same" as the left adjoint functor $\textbf{Psh} \to \textbf{Sh}$, if you handwave where the abelian group structure comes from. (To prove this rigorously you need the fact that $\textbf{Psh} \to \textbf{Sh}$ preserves finitary products.)

I think there are no reasonable conditions that ensure that the presheaf $\mathbb{Z} S$ is automatically a sheaf. For one thing, in the case $S = 1$, this amounts to asking for conditions for the constant presheaf $\mathbb{Z}$ to be a sheaf. This never happens when the site contains an empty object. I think a sufficient condition would be for the site to consist of only connected objects, i.e. no representable sheaf is initial, nor a coproduct of two or more representable sheaves.

Answer 2

More generally, suppose you are working in a topos with natural numbers object $\mathbb{N}$. (This is certainly the case for the category of sheaves on a site.) Then with the aid of the internal language of a topos, you can go through the construction of a free group in terms of formal products of elements of $A$ or their formal inverses:

• For an object $X$, you can encode $\operatorname{list}(X)$ as for example $\{ f : (X \sqcup \{ pastend \})^{\mathbb{N}} \mid (\forall n : \mathbb{N}, f(n) = pastend \rightarrow f(S(n)) = pastend) \land (\exists n : \mathbb{N}, f(n) = pastend) \}.$ It is also easy to define $cons$ and $nil$ operators in terms of this representation; and with only slightly more effort, you can define concatenation and reversal operators.
• Given a type of generators $A$, start off with $\operatorname{list}(A \sqcup A)$. For $x : A$, we will also denote $x$ (by slight abuse of notation) as the corresponding element in the left copy of $A$, and $\bar x$ as the corresponding element in the right copy of $A$.
• Now take the intersection of all equivalence relations on $\operatorname{list}(A\sqcup A)$ which satisfy $cons(x, cons(\bar x, \ell)) \equiv \ell$; $cons(\bar x, cons(x, \ell)) \equiv \ell$; and $\ell_1 \equiv \ell_2 \rightarrow cons(y, \ell_1) \equiv cons(y, \ell_2)$ for $y : A\sqcup A$. This again gives an equivalence relation which satisfies all these closure conditions.
• The quotient of $\operatorname{list}(A\sqcup A)$ by this equivalence relation will give the desired group to interpret as the free group on $A$. The identity is the equivalence class of the empty list; multiplication is induced by concatenation of lists; and inverse is induced by mapping each $x$ to $\bar x$ and vice versa, and then reversing the list.

Do note that as opposed to the case in $\mathbf{Set}$, we will not necessarily be able to reduce every such list to a normal form. The obstruction here is that to decide whether $x \bar y \cdots$ can be reduced, we would need to be able to decide whether $x = y$ as elements of $A$. (On the other hand, if $A$ does happen to have decidable equality, then you do indeed get the ability to define a normalization function on lists, and the free group on $A$ ends up being equivalent to the subtype of reduced lists.)

And then, once you have the free group on $A$, you can similarly proceed as usual to take the intersection of all normal subgroups containing all commutators, and take the quotient of the free group by the corresponding congruence to get the free abelian group on $A$. (As an alternative, you could have just included the commutativity requirement $cons(x, cons(y, \ell)) \equiv cons(y, cons(x, \ell))$ for $x, y : A\sqcup A$ in the equivalence relation construction above.)