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

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?

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 $$q$$. $$\square$$