**Edit.** What I originally wrote was wrong, because I was ignoring the (huge) K-theory of the ground field $k$. I corrected this below.

Let $k$ be an algebraically closed field. Let $X$ be a proper $k$-scheme of pure dimension $d$. Denote by $\pi$ the following unique $k$-morphism, $$\pi:X\to \text{Spec}\ k,$$ and denote by $i$ any $k$-point, $$i:\text{Spec}\ k \to X.$$ Denote the image of $i$ by $\widetilde{x}$, a specified base point. Because the higher Chow groups are covariant for proper morphisms, there are induced maps,
$$\text{CH}^{q}(i,n):\text{CH}^q(\text{Spec}\ k,n) \to \text{CH}^{d+q}(X,n),$$ $$\text{CH}^{d+q}(\pi,n):\text{CH}^{d+q}(X,n)\to \text{CH}^q(\text{Spec}\ k,n).$$
The composition $\text{CH}^{d+q}(\pi,n)\circ \text{CH}^q(i,n)$ is $\text{CH}^q(\text{Id}_{\text{Spec}\ k},n)$, which is the identity map.

**Definition.** A $k$-scheme $X$ is **strongly $\mathbb{A}^1$-connected** if for every pair $(x_0,x_1)$ in $X\times_{\text{Spec}\ k}X(k),$ there exists a $k$-morphism, $$u:(\mathbb{A}^1_k,0,1)\to (X,x_0,x_1).$$

**Proposition.** For every strongly $\mathbb{A}^1$-connected, proper $k$-scheme $X$ of dimension $d$, for every integer $n$, the morphisms $\text{CH}^{d+n}(\pi,n)$ and $\text{CH}^n(i,n)$ are inverse isomorphisms.

**Proof.** Via covariance, the composition $\text{CH}^{d+n}(\pi,n)\circ \text{CH}^n(i,n)$ equals the identity map. Thus, it suffices to prove that $\text{CH}^n(i,n)$ is surjective. Since $Z^{d+n}(X,n)$ has as $\mathbb{Z}$-basis the classes of $k$-points of $X\times \Delta^n$ that are in the complement of the face maps, it suffices to prove that every such class is cohomologous to the class of a $k$-point in the image of $i\times \text{Id}_{\Delta_n}$.

Let $(x,a)$ be a $k$-point of $X\times \Delta^n$ that is not in the image of the face maps, i.e.,

$$a=(a_0,\dots,a_{n-1},a_n)\in \Delta_{n} \setminus \cup_i \text{Image}(\delta^i_{n-1}), \ \ a_0+\dots+a_n=1, \ \ a_0\cdots a_n \neq 0.$$ For a the $k$-points $x$ and $\widetilde{x}$ of $X$, since $X$ is strongly $\mathbb{A}^1$-connected, there exists a morphism $u:(\mathbb{A}^1,0,1)\to (X,x,\widetilde{x})$. Now extend this to a morphism $$u:\mathbb{A}^1\to X\times \Delta_{n+1},\ \ \ t\mapsto (u(t),a_0,\dots,a_{n-1},a_nt,a_n(1-t)). $$ This is a closed immersion; in fact the composition with the projection to the last two factors is already a closed immersion. By construction, this does intersect transversally the faces of $X\times \Delta_{n+1}.$ Thus the class of the image is an element of $Z^{d+n}(X,n+1)$. Up to a sign, the differential of this class equals $[(x,a)]-[(\widetilde{x},a)].$ Thus, the class $[(x,a)]$ is cohomologous to the class $[(\widetilde{x},a)]$. Since $\widetilde{x}$ is the image of $i$, the class $[(\widetilde{x},a)]$ is in the image of $\text{CH}^n(i,n).$ **QED**

**Note.** For $n$ equal to $0,$ the Chow group $\text{CH}^0(\text{Spec}\ k,0)$ is isomorphic to $\mathbb{Z}$ via the degree homomorphism. Thus, for $n$ equal to $0$, the Proposition is Roitman's result. Based on the works of Yi Zhu, specifically the following, I also suspect that the argument above works for quasi-projective $k$-schemes $X$ that are strongly $\mathbb{A}^1$-connected, but where the higher Chow groups are replaced by Suslin's groups.

Yi Zhu.

$\mathbb{A}^1$-equivalence of zero cycles on surfaces.

To appear: Transactions of the AMS.

https://sites.google.com/site/yizhuhomepage/