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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 $\mathcal{O}_Y$.

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.

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 $\mathcal{O}_Y$.

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 $\mathcal{O}_Y$.

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 coduimension $1$.

This imples that $\textrm{Supp}(Q)$ has codimension at least $2$, hence $c_1(Q)=0$.

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

Hence $f^*(M) \otimes L^{-1} \in \textrm{Pic}^0(Y)$.

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

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 $\mathcal{O}_Y$.