Let $X$ be the variety $\mathbb{C}\backslash\{0\}$, and consider the map $f:X\to X$, $z\mapsto z^2$. I am trying to compute the induced map $f^*$ on the first étale cohomology of $X$ with coefficients in a finite abelian group. The answer I find is incorrect, and I wish to ask for help finding the error.

Let $n$ be an odd positive integer. Then $H^1_{et}(X,\mathbb{Z}/n)$ classifies $\mathbb{Z}/n$-torsors over $X$. Let $\zeta_n=e^{2\pi i/n}$, and consider the torsor $T=\mathbb{C}\backslash\{0\}$, with projection map $\pi:T\to X$ given by $\pi(z)= z^n$ and the generator of $\mathbb{Z}/n$ acting on $T$ by $z\mapsto\zeta_n z$. The pullback torsor $f^*T$ is the fiber product of $X$ and $T$ over $X$, with respect to the maps $f$ and $\pi$. I claim this fibre product is again $f^*T=\mathbb{C}\backslash\{0\}$, with maps to $X$ and $T$ given by $z\mapsto z^n$ and $z\mapsto z^2$, respectively. The map $f^*T\to T$ should intertwine the action of the generator of $\mathbb{Z}/n$ on $f^*T$ and $T$, so the action on $f^*T$ must be $z\mapsto \zeta_n^{(n+1)/2}z$. Since $\zeta_n^{(n+1)/2}$ is the unique square root of $\zeta_n$ in the group of $n$-th roots of unity, I conclude that $f^*$ acts by multiplication by $1/2$ on $H^1_{et}(X,\mathbb{Z}/n)$.

However, I know from topology that $f^*$ acts by multiplication by $2$ on the singular cohomology $H^1(X,\mathbb{Z}/n)$, and singular and étale cohomologies are supposed to agree over $\mathbb{C}$.

What is wrong with my computation of $f^*$ on $H^1_{et}(X,\mathbb{Z}/n)$?


Your calculations are correct, and the confusion just arises in how you're identifying torsors with cohomology classes. The $\check{\mathrm{C}}$ech $1$-cochains associated to $T$ and $f^*T$ can be explicitly written down in terms of $\mathbb{Z}/n\mathbb{Z}$-valued transition functions between local trivializations of $T$ and $f^*T$ on an open cover $\{U_i\}$ of $\mathbb{C}\setminus \{0\}$. Suppose the transition function computed for $T$ on some overlap $U_i \cap U_j$ is some element $a\in\mathbb{Z}/n\mathbb{Z}$, so that we're multiplying by $\zeta_n^a$. Picking the same trivializations for $f^* T$, the transition function on $U_i \cap U_j$ will also be multiplication by $\zeta_n^a$, but from the way we've defined the group action, the corresponding element in $\mathbb{Z}/n\mathbb{Z}$ is $2a$. Thus the cochain we compute for $f^*T$ will be twice the cochain computed for $T$.


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