Projection formula for invertible twists

Tensoring with an invertible sheaf $L$ is an exact autoequivalence, with inverse tensoring with $L^{\vee }$: exactness is checked in local trivializations. It preserves injectives, since $\mathrm{Hom}(M,I\otimes L)\cong \mathrm{Hom}(M\otimes L^{\vee },I)$ is exact in $M$ when $I$ is injective. The same statements hold on $X$ for $f^{\ast }L$.

For any module $G$ on $X$, adjunction gives the natural map ${\nu }_G:f_{\ast }G\otimes L\to f_{\ast }(G\otimes f^{\ast }L)$, adjoint to the counit map $f^{\ast }(f_{\ast }G)\otimes f^{\ast }L\to G\otimes f^{\ast }L$. On every open trivializing $L$ this is the identity under the trivializations, hence it is an isomorphism. This ordinary direct-image argument requires no quasi-compactness or separatedness of $f$.

Take an injective resolution $I^{\bullet }$ of $\mathcal{F}$. By step 1.1, $I^{\bullet }\otimes f^{\ast }L$ is an injective resolution of $\mathcal{F}\otimes f^{\ast }L$. Naturality of $\nu$ gives an isomorphism of complexes $f_{\ast }{I}^{\bullet }\otimes L\cong f_{\ast }(I^{\bullet }\otimes f^{\ast }L)$. Exact tensor with $L$ commutes with taking cohomology, so the resulting isomorphism is precisely $R^q{f}_{\ast }\mathcal{F}\otimes L\cong R^q{f}_{\ast }(\mathcal{F}\otimes f^{\ast }L)$ for every $q$.

If $f_{\ast }{\mathcal{O}}_X={\mathcal{O}}_Y$ and the higher direct images of ${\mathcal{O}}_X$ vanish, applying step 2.1 to ${\mathcal{O}}_X$ gives $f_{\ast }{f}^{\ast }L=L$ and $R^q{f}_{\ast }{f}^{\ast }L=0$ for $q>0$. For the $k$-morphism in the final assertion, the vanishing-direct-image comparison gives $k$-linear isomorphisms $H^n(X,f^{\ast }L)\cong H^n(Y,L)$. The proper schemes of the final assertion are locally Noetherian by [F9]; the line bundles are finite-type quasi-coherent modules by [F3], hence coherent by [F9], and their cohomology is finite-dimensional and vanishes in sufficiently high degree. Taking the finite alternating sums proves the Euler-characteristic identity.