In these notes the following theorem is stated, among other things.

Let $X$ be a pointed connected geometrically unibranch scheme over $\mathbb{C}$. Then Artin-Mazur etale homotopy type of $X$ is equivalent to the profinite completion of the homotopy type of the topological space $X(\mathbb{C})$.

Artin has shown in SGA 4 Expose XVI that for a scheme $Y$ of finite type over $\mathbb{C}$ and the constant sheaf $F$ associated to a finite abelian group we have an isomorphism $$ H^n_{et}(Y, F)=H^n(Y(\mathbb{C}), F). $$

The second theorem is, in some sense, stronger than the first as it is not restricted to unibranch schemes. Is it possible to give a stronger version of the first theorem which would describe the etale homotopy type (and not only the cohomology groups) for non-unibranch schemes?


1 Answer 1


If $X$ is not geometrically unibranch, the etale homotopy type might not be profinite anymore, but Artin and Mazur proved (in Chapter 12, Theorem 12.9) that its profinite completion agrees with the profinite completion of $X(\mathbb{C})$.

Maybe this is not the best possible result, but it's not clear how to improve it. Think about the curve $X$ obtained by identifying two points on $\mathbf{G}_m$. Then $X(\mathbb{C})$ has the homotopy type of $S^1 \vee S^1$, so $\pi_1(X(\mathbb{C}))$ is the free product of $\pi_1(\mathbb{G}_m(\mathbb{C}))\simeq \mathbb{Z}$ and $\pi_1(\text{loop through node}) \simeq \mathbb{Z}$. For $\pi_1^{\rm SGA3}(X)$, which is I think the $\pi_1$ of the etale homotopy type, we should get the free product of $\pi_1(\mathbb{G}_m)\simeq \widehat{\mathbb{Z}}$ and $\pi_1(\mathbb{A}^1/1\sim 0)\simeq \mathbb{Z}$. How should the homotopy type of $X(\mathbb{C})$ know which loop generates a discrete copy of $\mathbb{Z}$ inside $\pi_1$ of the etale homotopy type?

This seems to indicate that without the geom. unibranch assumption, the homotopy type of $X(\mathbb{C})$ does not determine the etale homotopy type of $X$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.