Local fibration vs. stalkwise fibration

Question

Let $\mathbf{C}$ be a Grothendieck site with enough points. Let $p:\mathcal{E}\to \mathcal{F}$ be a map of simplicial presheaves on $\mathbf{C}$. Is it true that $p$ is a local (Kan) fibration if and only if it is a stalkwise fibration?

To fix terminology:

$p$ is called stalkwise fibration if for each point $q$ the map $q^*(\mathcal{E})\to q^*(\mathcal{B})$ is a fibration.

$p$ is called local fibration if for each $U\in \mathbf{C}$ and each commutative diagram

$$\require{AMScd} \begin{CD} \Lambda^n_k @>>> \mathcal{E}(U)\\ @VVV @VVV \\ \Delta^n @>>> \mathcal{B}(U) \end{CD}$$

there exists a covering family $(V_i\to U)_i$ such that in the induced square

$$\require{AMScd} \begin{CD} \Lambda^n_k @>>> \Pi_i\mathcal{E}(V_i)\\ @VVV @VVV \\ \Delta^n @>>> \Pi_i\mathcal{B}(V_i) \end{CD}$$

has a lifting $\Delta^n\to \Pi_i\mathcal{E}(V_i)$.

Answer 1

These are equivalent.

If $K$ is a simplicial set, and $\mathcal{F}$ is a simplicial presheaf, then there's a presheaf of sets $\mathcal{F}^K$, defined by $(\mathcal{F}^K)(U) = \hom(K, \mathcal{F}(U))$, where $\hom$ is maps of simplicial sets.

The important observation here is that if $K$ is a finite simplicial set, then formation of this gadget commutes with sheafification: $q^*(\mathcal{F}^K)\approx (q^*\mathcal{F})^K$. This is because $\mathcal{F} \mapsto \mathcal{F}^K$ is computed as a finite limit, if $K$ is finite.

Now consider the map of presheaves of sets $f: \mathcal{E}^{\Delta^n} \to \mathcal{E}^{\Lambda^n_k}\times_{\mathcal{B}^{\Lambda^n_k}} \mathcal{B}^{\Delta^n}$. Your map $p$ is a local fibration if the sheafification of $f$ is an epimorphism; the map $p$ is a stalkwise fibration if $q^*(f)$ is a surjection for each point $q$. If you have enough points, these mean the same thing.

(This is addressed in the introduction to the paper by Jardine, "Boolean localization in practice" (Documenta Mathematica, v.1), where he tells you what to do even if you don't have enough points!)