Question

Let $k$ be a field and $X$ a smooth $k$-scheme. We consider now the pointed (constant) simplical Nisnevich-sheaf $X_{+}$ that is represented by $X$. Let now $\nu\in U$ be a point of a smooth scheme, then the fibre of $X_{+}$ at $\nu$ is defined as $\text{colim} \ X_{+}(V)$ where the colimit is over all Nisnevich neighborhoods $V$ of $\nu$. What is this fiber now? Is this just the constant simplicial set associated to the set of all points of $X$ with residue field isomorphic to the one of $\nu$?

And as a further question; If I have a morphism $f:X\rightarrow Y$ between two smooths $k$-schemes, when is the associated morphism of simplicial Nisnevich sheaves $f_{+}:X_{+}\rightarrow Y_{+}$ a simplicial weak equivalence? For example in the case where $Y$ is irreducible and $f$ an open immersion of a dense open subscheme $X$?

Answer 1

Let $N:=\textrm{Neigh}_{Nis}(U,\nu)$ denote the filtered category of Nisnevich neighborhoods of the point $\nu\in U$. Then $\textrm{colim}_{V\in N^{op}} V\cong \textrm{spec}(\mathcal{O}^{h}_{U,\nu})$, where $N^{op}$ denotes the opposite category of $N$ and $\mathcal{O}^{h}_{U,\nu}$ is the henselization of the local ring $\mathcal{O}_{U,\nu}$. My guess now would be that the fiber is the constant simplicial set associated to the set $\textrm{Hom}_{k}(\textrm{spec}(\mathcal{O}^{h}_{U,\nu}),X)$.

Maybe someone can clarify this as well as your second question.

Answer 2

It looks like you probably need a stronger condition for the lements of the fiber. Specifically, it is not enough that residue fields should be isomorphic. There also must be a morphism on the Nisnevich local rings. Thus, I imagine that $$colim X_+(V)$$ should correspond to the set of all points of $X$ such that the henselization of the local ring is isomorphic to the henselization of the local ring of $\nu$.