**Situation:** Let $\Bbbk$ be an algebraically closed field. Assume that $\pi:Y\to X$ is an finite, dominant, *unramified* morphism between nonsingular varieties of dimensions $n$. Let $d=\deg(\pi)$. 


**What I know:** For $\Bbbk=\mathbb C$, the second chern class of $X$ equals its topological Euler characteristic (i.e., the Euler characteristic with respect to the topology of a complex manifold). I know this under the name *Gauss-Bonnet Formula*. It then follows that $c\_n(Y)=d\cdot c\_n(X)$ because $\pi$ is a $d$-fold covering map of complex manifolds. 


**My Question:** Does $c\_n(Y)=d\cdot c\_n(X)$ hold under the more general assumption that $\Bbbk$ is algebraically closed? In particular, does this hold in positive characteristic? 


PS: I am mostly interested in the case $n=2$, i.e. $\pi$ is a covering map of surfaces. However, I felt that this would probably work for any $n$.



Top chern class under finite, unramified, dominant morphism
Yes, it does hold in positive characteristic. You can show that the degree of $c\_n(X)$ equals its Euler characteristic with respect to étale cohomology with coefficients in $\mathbb{Q}\_{\ell}$, where $\ell$ is a prime different from the characteristic ( [Top chern class in positive characteristic](https://mathoverflow.net/questions/80697/top-chern-class-in-positive-characteristic) ), and this is has the required behavior under étale covers ( [Behaviour of euler characteristics in characteristic p for finite etale covers](https://mathoverflow.net/questions/41576/behaviour-of-euler-characteristics-in-characteristic-p-for-finite-etale-covers) ).


[Edit:] the equality can be proved directly; in fact the proof is much easier. Since $\pi\colon Y\to X$ is étale, the tangent bundle $T\_Y$ is the pullback $\pi^\*T\_X$. By functoriality of Chern classes, $c\_n(Y) = \pi^\*c\_n(X)$ (at the level of Chow groups). But it easy to see that the composite $\pi\_\*\pi^\*$ from the Chow group of $X$ to itself is just multiplication by $d$.


