Line bundles and rational singularities

Question

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

Answer 1

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.

Answer 2

Since $X$ has rational singularities, it is normal. Then the singular locus of $X$ has codimension at most $2$ and $f \colon X \to Y$ is an isomorphism in codimension $1$.

This imples that the support of $Q$ has codimension at least $2$, hence $c_1(Q)=0$.

Therefore your exact sequences gives $c_1(f^*M)=c_1(L)$, that is $c_1(f^*M \otimes L^{-1})=0$, that is $f^*M \otimes L^{-1} \in \textrm{Pic}^0(Y)$.

Now $H^0(Y, f^*M \otimes L^{-1})=H^0(Y, \mathcal{O}_Y)=\mathbb{C}$ implies that $f^*M \otimes L^{-1}$ is a line bundle of degree $0$ with a non-zero global section, hence it must be isomorphic to $\mathcal{O}_Y$.

This implies $Q=0$: in fact, we obtain $$\textrm{Hom}(f^*M, L)=H^0(Y, f^*M^{-1} \otimes L)=H^0(Y, \mathcal{O}_Y)=\mathbb{C},$$ so $\eta\colon f^*M \longrightarrow L$ is necessarily an isomorphism.