# Does the functor $\mathbb{Z}_*: PSh(\mathbf{C})\to PSh^{Ab}(\mathbf{C})$ send sheaves to sheaves?

Let $\mathbf{C}$ be a Grothendieck site. Does the functor $$\mathbb{Z}_*: PSh(\mathbf{C})\to PSh^{Ab}(\mathbf{C})$$ from presheaves of sets $Fun(\mathbf{C}^{op}, Set)$ to preasheaves of abelian groups $Fun(\mathbf{C}^{op}, Ab)$ defined by the free abelian group on every section send sheaves to sheaves?

I doubt this as $\mathbb{Z}:Set\to Ab$ is a left adjoint and the sheaf condition is a limit.

## 1 Answer

Let $X = \{ P, Q \}$ be the discrete two-point space. A presheaf $F$ on $X$ is a sheaf if and only if $F(X) \cong F(P) \times F(Q)$ and $F(\varnothing)$ is the initial object.

Thus, if $F$ is a sheaf, we have

$$\mathbf{Z}_*F(X) \cong \mathbf{Z}[F(P) \times F(Q)] \cong \mathbf{Z}[F(P)] \otimes \mathbf{Z}[F(Q)] \cong \mathbf{Z}_*F(P) \otimes \mathbf{Z}_*F(Q)$$

If, for example, $F(P)$ and $F(Q)$ are both three point sets, then $\mathbf{Z}_*F(P)$ and $\mathbf{Z}_*F(Q)$ are free abelian groups on three elements, and their product is a free abelian group on six elements. Howver, $\mathbf{Z}_*F(X)$ is a free abelian group on nine elements.

It turns out that even when $F(P)= \{0,1\}$ and $F(Q)=\{2,3\}$, it's still not a sheaf; e.g. the elements $[(0,2)] + [(1,3)]$ and $[(0,3)] + [(1,2)]$ are distinct elements of $\mathbf{Z}_*F(X)$, but they have the same restriction to $\mathbf{Z}_*F(P)$ and to $\mathbf{Z}_*F(Q)$.