The Stacks project

The meaning of the statement is not completely formal. To make sense of "the identity on cohomology" we need to show that we can identify $\mathcal{F}$ with $g_*(\mathcal{F})$ and/or $g^{-1}(\mathcal{F})$, for any $\mathcal{F}$. This is of course the case in subsequent lemmas where $\mathcal{F}$ is a constant sheaf.

Hmm... yes. OK, well, I think Exercise 64.3.1 about topological spaces just above the lemma explains why this would be so. To be precise $g_*\mathcal{F}(U) = \mathcal{F}(U \times_{\varphi, X, g} X)$ which is functorially equal to $\mathcal{F}(U)$ by the assumption of the lemma. So indeed $\mathcal{F} = g_*\mathcal{F}$ for all sheaves $\mathcal{F}$ and hence also $g^{-1}\mathcal{F} = \mathcal{F}$ for all sheaves $\mathcal{F}$. Then of course this agrees with what you would do on constant sheaves. Sigh!