Let $f:X\to Y$ be a finite cover of smooth algebraic varieties, branched along a divisor $R\subset Y$. Let $E$ be a vector bundle on $Y$. What is the relation between the chern classes of $E$ and the chern classes of $F:=f\_\*(f^\*E)$?



chern classes of push-pulled vector bundles
You need to know what is the splitting of $f\_\* \mathscr{O}\_X$, which depends on the cover.


Let us consider the double cover case, which is the easiest one. Then $$f\_\* \mathscr{O}\_X = \mathscr{O}\_Y \oplus L^{-1},$$ where $L$ is a line bundle on $Y$ such that $R \in |2L|$. Then, given any vector bundle $E$ on $Y$, by using projection formula one obtains
$$F=f\_\*(f^\*E)=f\_\*(f^\*E \otimes \mathscr{O}\_X) = E \otimes f\_\* \mathscr{O}\_X = E \oplus (E \otimes L^{-1}),$$
so the computation of the Chern classes of $F$ is now a straightforward application of the splitting principle.


For general finite covers of degree $d$, one can only say that $$f\_\* \mathscr{O}\_X = \mathscr{O}\_Y \oplus V,$$
where $V$ is a vector bundle on $Y$ of rank $d-1$. As before, one obtains $$F=f\_\*(f^\*E)=f\_\*(f^\*E \otimes \mathscr{O}\_X) = E \otimes f\_\* \mathscr{O}\_X = E \oplus (E \otimes V),$$
but now the computations of Chern classes will be more difficult since in general $V$ is indecomposable. 


In some lucky cases, for instance when the cover $f \colon X \to Y$ is Galois with abelian Galois group, $V$ splits completely into line bundles and then, if one is able to identify $V$, the computation can be made again by using the splitting principle.


