Hi, I have some problem to understand the proof of lemma 3.2 of this article: <http://www.ams.org/journals/jams/2001-14-03/S0894-0347-01-00368-X/>.


The lemma states the following:
Let $X$ be a variety and $f: Y \rightarrow X$ a resolution of singularities. Assume that $X$ has rational singularities. Then a line bundle $L$ on $Y$ is the pullback $f^\*M$ of some linebundle $M$ on $X$ if and only if the restriction of $L$ to each formal fibre of $f$ is trivial. Moreover, when this holds, $M=f\_\*L$.


For the proof of the "if" part, suppose that the restriction of $L$ to each formal fibre is trivial. The teorem on formal functions shows that the completions of the stalks of the sheaves $R^if\_\* \mathcal{O}\_Y$ and $R^if\_\*L$ at any point $x \in X$ are isomorphic for each $i$. Since $X$ has rational singularities, $R^if\_\*L=0$ for all $i>0$ and $M=f\_\*L$ is a linebundle on $X$.


Since $f^\*M$ is torsion free, the natural adjunction map $\eta: f^\*f\_\*L \rightarrow L$ is injective, so there is a short exat sequence 
$$ 0 \rightarrow f^\*f\_\*L \stackrel{\eta}{\rightarrow} L \rightarrow Q \rightarrow 0.$$
By tha projection formula and the fact that $X$ has rational singularities, $R^if\_\*(f^\*M)=M \otimes R^if\_\* \mathcal{O}\_Y=0$ for all $i>0$. The fact that $\eta$ is the unit of adjunction implies that $f\_\* \eta$ has a left inverse, and in particular is surjective. Applying $f\_\*$ to the exact triple we conclude that $f\_\*Q=0$, and, by the theorem on formal functions $f\_\*(Q \otimes L^{-1})=0$, in particular $Q \otimes L^{-1}$ has no nonzero global sections. Tensoring the exact triple with $L^{-1}$ gives a contradiction, unless $Q=0$. Hence $\eta$ is an isomorphism and we are done.


I did not understand this last step. Tensoring the short exact sequence with $L^{-1}$ and then taking global sections, we get $ \Gamma(Y, f^\*M \otimes L^{-1}) \cong \Gamma(Y, \mathcal{O}\_Y)$, because the last term is zero. How can I deduce from this that $f^\*M \otimes L^{-1} \simeq \mathcal{O}\_Y$ and then $f^\*M \cong L$? Where is the contradiction? Thank you



Line bundles and rational singularities
For any line bundle $F$, giving a map of sheaves $\mathcal O\_X \to F$ is equivalent to giving a global section of $F$.


In your case, take $F=Q \otimes L^{-1}$ with the map given by your exact sequence tensorized by $L ^{-1}$. As $F$ has no non-zero global section, the aforesaid map is trivial, so $\eta \otimes Id\_{L^{-1}}$ is an isomorphism, and therefore so is $\eta$ too. 


