$f_\ast\mathcal{O}_X$ is locally free of rank $r=\deg f$ for a finite $r$-to-one morphism $f$. Let $f\colon X\to Y$ be a finite $r$-to-one morphism between smooth projective varieties.
How does one show that $f_\ast\mathcal{O}_X$ is a locally free sheaf of rank $r$ on $Y$ ?
Any insights for $f_\ast\mathcal{O}_X$ would be very much appreciated. A local section over $U\subset Y$ for $f_\ast\mathcal{O}_X$ is just a regular function over $f^{-1}(U)$, so it seems as if one can write this uniquely as a sum of $r$ "pieces"...?
An aside second question: is there a naturally defined map $f^{\ast}f_\ast\mathcal{O}_X(D)\to\mathcal{O}_X(D)$ where $D$ is a divisor on $X$ ?
 A: The key non-trivial ingredient is that $f$ is automatically flat in this situation. This is shown in 'Commutative Ring Theory' by Matsumura in Theorem 23.1, which states:
If $A \subset B$ are local noetherian regular domains (We can even assume $B$ to be only Cohen-Macaulay, but we do not need this in your case), $\mathfrak m \subset A$ the maximal ideal of $A$ and $\dim A = \dim B + \dim (B \otimes_A A/\mathfrak m)$ holds, then $B$ is flat over $A$.
If $A \subset B$ is finite, the equality of dimensions is automatically satisfied, since $A/\mathfrak m \subset B/\mathfrak m B$ is also finite, hence $B/\mathfrak m B$ is $0$-dimensional. So $B$ is $A$-flat if $A \subset B$ is a finite extension of local noetherian regular domains.

Once we know that $f$ is flat, it is easy: Everything is noetherian, so any finitely generated module is automatically finitely presented, hence finite and flat implies projective, which implies locally free. Finding the rank $m$ is also easy, let $U = \operatorname{Spec}(A) \subset Y$ an affine open, that trivializes $f_*\mathcal O_X$. Then - since $f$ is affine - $f^{-1}(U)$ is also affine, say $f^{-1}(U) = \operatorname{Spec}(B)$.
We have $B = A^{\oplus m}$ and passing to the quotient fields yields $k(X) = k(Y)^{\oplus m}$, i.e. $m=[k(X):k(Y)]$ is the degree of the map $f$.
