Our goal is to prove the following theorem. Let $k$ be a field.
Theorem 1. Let $\mathscr{F} \in \mathsf{PST}(k)$ that is homotopy invariant. For $X \in \mathsf{Sm}_ k$ we assume that $\mathscr{F}(X)$ is torsion with exponent prime to $\operatorname{char} k$. Then $\mathscr{F}_ \mathrm{et}$ is locally constant.
Here, what is locally constant? Recall that there is a functor $\mathsf{Et}_ k \to \mathsf{Sm}_ k$ that induces adjoint functors $$ \pi^\ast \colon \mathsf{Sh}(\mathsf{Et}_ k) \leftrightarrow \mathsf{Sh}(\mathsf{Sm}_ k) \colon \pi_\ast. $$ Locally constant then means that it looks like $\pi^\ast$ of a $\Gal(k^\mathrm{sep}/k)$-module, i.e., $\pi^\ast \pi_\ast \mathscr{F} \cong \mathscr{F}$.
So let’s prove the theorem. We now want to check the isomorphism $\pi^\ast \pi_\ast \mathscr{F} \cong \mathscr{F}$ on stalks. We can reduce to the case when $k = k^\mathrm{sep}$, and so we just need to show that $$ \mathscr{F}(\mathcal{O}_ {\mathbb{A}^n,0}^\mathrm{sh}) \cong \mathscr{F}(\Spec k). $$ Here, we have $$ \Spec k = S_0 \subseteq S_1 \subseteq \dotsb \subseteq S_n, $$ where $S_e$ is the strict henselization of $0 \in \mathbb{A}^e$. Note that there are also splittings $S_{e+1} \to S_e$ because we have $\mathbb{A}^{e+1} \to \mathbb{A}^e$. We already know that there are maps $$ i_e^\ast \colon \mathscr{F}(S_{e+1}) \twoheadrightarrow \mathscr{F}(S_e), $$ which are surjective due to these splittings, it remains to show that they are injective.
More concretely, given any element of $\mathscr{F}(S_e)$, it comes from some $$ S_e \xrightarrow{\pi} X \to \mathbb{A}^e, $$ where $(X, x_0) \to (\mathbb{A}^e, 0)$ is pointed and étale. So given any $\pi^\ast \varphi \in \mathscr{F}(S_e)$ such that $i_e^\ast \pi^\ast \varphi = 0$, we want to show that $\pi^\ast \varphi = 0$.
Because $X \to \mathbb{A}^e$ is étale, its image $U \subseteq \mathbb{A}^e$ is open and choosing a linear map we have $$ X \to \mathbb{A}^e \to \mathbb{A}^{e-1} $$ that is a smooth relative curve. Now we find a good compactification $$ X \hookrightarrow \bar{X} \to \mathbb{A}^{e-1} $$ where $\bar{X}$ is normal and $\bar{X} - X$ admits an affine neighborhood. Such a compactification can be constructed for example by applying Zariski main theorem to get a decomposition $$ X \hookrightarrow \bar{X} \xrightarrow{\text{finite}} \mathbb{A}^{e-1} \times \mathbb{P}^1. $$
We now consider the fiber product $$ X^\prime = X \times_{\mathbb{A}^{e-1}} S_e \xrightarrow{q} X, $$ where the map is $S_e \to S_{e-1} \to \mathbb{A}^{e-1}$. There are two maps $$ \pi, \pi \circ i_e \circ \pi_e \colon S_e \to X, $$ and these induce two different maps $$ s, s^\prime \colon S_e \to X^\prime. $$ We want to show that if $(\pi \circ i_e \circ \pi_e)^\ast \varphi = 0$ then $\pi^\ast \varphi = 0$. Here, we see that $$ (\pi \circ i_e \circ \pi_e)^\ast \varphi = (s^\prime)^\ast (q^\ast \varphi), \quad \pi^\ast \varphi = s^\ast (q^\ast \varphi). $$ So it suffices to prove the following.
Claim 2. For any $\psi \in \mathscr{F}(x^\prime)$ we have $s^\ast \psi = (s^\prime)^\ast \psi$ if $s_0^\ast \psi = (s_0^\prime)^\ast \psi$, where $s_0$ and $s_0^\prime$ are the special fibers over $0 \in S_e$.
For $X \to S$ smooth, we can define $C_0(X/S)$ to be the free abelian group generated by irreducible closed subschemes $Z \subseteq X$ that are finite surjective over $S$.
Claim 3. If $\mathscr{F}$ is homotopy invariant, we have a factorization $$ C_0(X^\prime \times_k S_e / S_e) \otimes \mathscr{F}(X^\prime) \twoheadrightarrow H_0^\mathrm{sing}(X^\prime \times_k S_e / E_e) \otimes \mathscr{F}(X^\prime) \to \mathscr{F}(S_e) $$ of the evaluation map.
What is the singular cohomology? If we unravel the definition, we see that $$ H_0^\mathrm{sing}(X/S) = \coker(C_0(X \times \mathbb{A}^1 / S \times \mathbb{A}^1) \xrightarrow{\partial_0 - \partial_1} C_0(X / S)). $$
Now we have a diagram $$ \begin{CD} C_0(X^\prime \times_k S_e / S_e) \otimes \mathscr{F}(X^\prime) @>>> H_0^\mathrm{sing}(X^\prime \times_k S_e / S_e) \otimes \mathscr{F}(X^\prime) @>>> \mathscr{F}(S_e) \br @VVV @VVV @VVV \br C_0(X_0 \times_k S_0 / S_0) \otimes \mathscr{F}(X^\prime) @>>> H_0^\mathrm{sing}(X_0/S_0) \otimes \mathscr{F}(X^\prime) @>>> \mathscr{F}(S_0). \end{CD} $$ We also see that $(\Gamma_s - \Gamma_{s^\prime}) \otimes \psi$ on the top left corner maps to $s^\ast \psi - (s^\prime)^\ast \psi$ on the right upper corner, and $s_0^\ast \psi - (s_0^\prime)^\ast \psi$ on the lower right corner. So it suffices to show that the middle map $$ H_0^\mathrm{sing}(X^\prime \times_k S_e/S_e) \otimes \mathscr{F}(X^\prime) \to H_0^\mathrm{sing}(X_0 / S_0) \otimes \mathscr{F}(X^\prime) $$ is injective.
At this point, we need to use our assumption. We note that we have $$ \mathscr{F}(X) = \bigcup_{(n,\operatorname{char} k) = 1} \mathscr{F}(X)[n]. $$ So it is enough to show that $$ H_0^\mathrm{sing}(X^\prime \times S_e / S_e) / n \cong H_0^\mathrm{sing}(X / S_0) / n. $$
Recall we have a good compactification $X \hookrightarrow \bar{X}$ with complement $Y$. Then we have the sheaf $$ \mathbb{G}_ {m,\bar{X},Y} = \ker(\mathscr{O}_ \bar{X}^\times \to i_{Y\ast} \mathscr{O}_ Y^\times), $$ and we consider $$ \operatorname{Pic}(\bar{X}, Y) = H^1(\bar{X}, \mathbb{G}_ {m,\bar{X},Y}) = \lbrace (\mathscr{L}, t) : \mathscr{L} \in \operatorname{Pic}(\bar{X}), t \colon \mathscr{L} \vert_Y \cong \mathscr{O}_ Y \rbrace. $$ This naturally fits in a short exact sequence $$ \mathscr{O}^\times(\bar{X}) \to \mathscr{O}^\times(Y) \to H^1(\bar{X}, \mathbb{G}_ {m,\bar{X},Y}) \to H^1(\bar{X}, \mathscr{O}_ \bar{X}^\times) \to H^1(Y, \mathscr{O}_ Y^\times). $$ The four terms on the outside are all homotopy invariant, so by the five lemma we also have $$ \operatorname{Pic}(\bar{X} \times \mathbb{A}^1, Y \times \mathbb{A}^1) \cong \operatorname{Pic}(\bar{X}, Y). $$
On the other hand, $n$ is invertible in $k$, so we have the Kummer exact sequence $$ 0 \to j_! \mu_n \to \mathbb{G}_ {m,\bar{X},Y} \xrightarrow{n} \mathbb{G}_ {m,\bar{X},Y} \to 0. $$ This gives $$ \operatorname{Pic}(\bar{X}, Y) \xrightarrow{n} \operatorname{Pic}(\bar{X}, Y) \to H^2(\bar{X}, j_! \mu_n) = H_c^2(X, \mu_n). $$
Theorem 4. If $X$ is smooth and $p \colon X \to S$ is a smooth quasi-affine relative curve with good compactification, then $H_0^\mathrm{sing}(X/S) \cong \operatorname{Pic}(\bar{X}, Y)$.
There is at least a map $C_0(X/S) \to \operatorname{Pic}(\bar{X},Y)$ because a $Z \subseteq X$ defines an effective divisor. This is surjective because the boundary is contained in an affine neighborhood. For injectivity, we note that any trivial line bundle corresponds to a principal divisor corresponding to $f \in K(\bar{X})$, and we check that this can be chosen so that $tf + (1-f)$ is well-defined.
Now combining these results we see that $$ H_0^\mathrm{sing}(X/S) / n \hookrightarrow H_0^\mathrm{sing}(X_0/S_0) / n $$ is injective, because both sides embed in $H_c^2(X, \mu_n)$.