Is the $\mathbb A^1$ homotopy type of a punctured curve independent of the choice of puncture?

Question

Let $C$ be a smooth connected algebraic curve over $\mathbb C$. Assume that $C$ admits no automorphisms. Let $p_1, p_2 \in C$ be two distinct points.

Question: Are the $\mathbb A^1$ homotopy types of $C - p_1$ and $C- p_2$ the same?

I think that the answer is "no", but I do not know how to construct an invariant that distinguishes the two. Of course, the ordinary homotopy types of $C-p_1$ and $C-p_2$ are the same. I also think that the stable $\mathbb A^1$ homotopy types are the same because of the purity cofiber sequence $$C-p_{i} \to C \to (\mathbb A^1, \mathbb A^1 - 0).$$ Is there a way of seeing that these two are not $\mathbb A^1$ homotopic, or of constructing an equivalence between them?

Answer 1

If I understand it correctly, this follows from the results of Severitt's master's thesis. (I am not an expert on this, so it is possible I misread one of the statements.)

Indeed, Lemma 9.1.1 shows that smooth projective curves of genus $> 0$ are $\mathbf A^1$-rigid. But the proof actually shows the following:

Lemma. Let $X$ be a smooth variety such that every map $\mathbf A^1 \to X$ is constant. Then $X$ is $\mathbf A^1$-rigid. $\square$

Now if $X = C - p$ for some smooth projective curve $C$ with $\operatorname{Aut}(C) = 1$, then in particular $g(C) > 0$ so every map $\mathbf A^1 \to C$ (or $\mathbf A^1 \to C \setminus p$) is constant. Thus, the lemma implies $X$ is $\mathbf A^1$-rigid. Then Theorem 6.3.7 implies the result: if $C - p$ and $C - q$ are (weakly) $\mathbf A^1$-homotopy equivalent, then the theorem gives $C - p \cong C - q$, which implies that there is a nontrivial automorphism of $C$ taking $p$ to $q$. $\square$