Let $\mu:Y\to X$ be a birational morhism between normal projective (complex) varieties.
Suppose, furthermore, that $Y$ is smooth. Let $D$ be a Weil divisor on $X$ and let $\mathcal{O}\_X(D)$ be its corresponding reflexive sheaf.
If we denote by $\widetilde{D}$ the strict transform of $D$, is it true that
the push-forward of the invetible sheaf $O\_Y(\widetilde{D})$ is equal to $\mathcal{O}\_X(D)$?



push-forward and strict transforms
No, the is definitely not true. Let me give you an example, suppose that $X = \mathbb{A}^2$ and that $\pi : Y \to X$ is the blowup of $X$ at the origin with exceptional divisor $E$. Suppose that $D$ is the Cartier divisor corresponding made up of $n$ lines through the origin, $n \geq 2$. Then $\widetilde{D}$ is just a disjoint union of $n$ lines on the blown up space, and $\pi^\* D = \widetilde{D} + nE$.


Suppose now that $\pi\_\* O\_Y(\widetilde D) = O\_X(D)$. Then by the projection formula, 
$$
\pi\_\* O\_Y(-nE) = \pi\_\* O\_Y(\widetilde D - \pi^\* D) = O\_X
$$
However, it is obvious that $\pi\_\* O\_Y(-nE) = \mathfrak{m}^n$ is the maximal ideal of the origin raised to the $n$th power. This is a contradiction.


