I am confused about the following (probably obvious) assertion.
Let $X, Y$ be topological spaces and let $f: X \to Y$ be a continuous map. Let $\mathcal{F}$ be a sheaf of abelian groups on $Y$ and let $f^{-1}Y$ be the inverse image.
In a book I am reading, it is claimed without proof that there is a natural map $R \Gamma(F) \to R\Gamma(f^{-1} F)$, where $\Gamma(-)$ is the global sections functor.
Why does this map exist? (I agree that there is a natural map $\Gamma(F) \to \lim_{V \supset f(f^{-1}(U)} F(V) \to \Gamma(f^{-1} F)$, but I don't see why we have induced maps at the level of derived functors.)
Let $\mathcal{F} \to \mathcal{J}^{\bullet}$ be an injective resolution in $\mathbf{Ab}(Y)$; then $f^{-1}$ gives a resolution $f^{-1}\mathcal{F} \to f^{-1}\mathcal{J}^{\bullet}$ (the $f^{-1}\mathcal{J}^{n}$ may not be injective). Let $f^{-1}\mathcal{F} \to \mathcal{I}^{\bullet}$ be an injective resolution in $\mathbf{Ab}(X)$. By "Comparison Theorem 2.3.7" of Weibel, Homological Algebra, there exists a morphism of complexes $f^{-1}\mathcal{J}^{\bullet} \to \mathcal{I}^{\bullet}$ extending the identity $f^{-1}\mathcal{F} \to f^{-1}\mathcal{F}$. Taking global sections $\Gamma(X,-)$ gives a morphism of complexes of abelian groups $\Gamma(X,f^{-1}\mathcal{J}^{\bullet}) \to \Gamma(X,\mathcal{I}^{\bullet})$. We precompose by the natural morphisms $\Gamma(Y,\mathcal{J}^{\bullet}) \to \Gamma(X,f^{-1}\mathcal{J}^{\bullet})$ in your post.