Does $f_*f^*(L)\cong L$ hold for a birational morphism?

Question

Suppose $X\to Y$ be a birational morphism of non-singular varieties, do we have $f_*f^*(L)\cong L$ for $L$ is a quasi-coherent sheaf on $Y$? Especially when $L$ is a line bundle?

Answer 1

The projection formula says that $f_*(\mathcal F \otimes f^*\mathcal E) = (f_*\mathcal F) \otimes \mathcal E$ for any coherent sheaf $\mathcal F$ on $X$ and any locally free sheaf $\mathcal E$ on $Y$.

In particular, $f_*f^*L = (f_*\mathcal O_X)\otimes L.$ So your question amounts to asking if $f_*\mathcal O_X = \mathcal O_Y$ (the special case of the trivial line bundle); this will then give the general case by the preceding formula.

Now if $f$ is not proper, there is not really much chance that $f_*\mathcal O_X = \mathcal O_Y$. (Consider the case when $f$ is affine, for example.)

But if $f$ is proper, as well as birational, then $f_*\mathcal O_X =\mathcal O_Y$ provided that $Y$ is normal (so certainly non-singular is okay). The point is that $f_*\mathcal O_X$ is a coherent $\mathcal O_Y$-algebra (by properness) and generically it coincides with $\mathcal O_Y$ (by birationality). Now normality kicks in to give the desired equality.