The stalk of an inverse image sheaf is the stalk over the image point

By [F1] and [L1], it is enough to identify the stalk $(f_{\mathrm{p}}\mathcal{G}{)}_x$.

If $t\in \mathcal{G}(V)$ with $f(x)\in V$, then $x\in f^{-1}(V)$ and $t$ defines a class $[V,t]\in (f_{\mathrm{p}}\mathcal{G})(f^{-1}(V))$. Sending the germ $t_{f(x)}$ to the germ of $[V,t]$ at $x$ gives a map $\mathrm{\Phi }:{\mathcal{G}}_{f(x)}⟶(f_{\mathrm{p}}\mathcal{G}{)}_x\mathrm{.}$ If two representatives agree on a smaller neighbourhood of $f(x)$, then their induced sections agree on the inverse image of that smaller neighbourhood, so $\mathrm{\Phi }$ is well defined.

Conversely, represent a germ in $(f_{\mathrm{p}}\mathcal{G}{)}_x$ by a section $[V,t]\in (f_{\mathrm{p}}\mathcal{G})(U)$ with $x\in U$ and $f(U)\subseteq V$. Since $f(x)\in V$, the section $t$ has a germ $t_{f(x)}\in {\mathcal{G}}_{f(x)}$. This depends only on the original germ at $x$, because equality of germs in $(f_{\mathrm{p}}\mathcal{G}{)}_x$ means equality after restricting to some smaller neighbourhood of $x$, hence after restricting $t$ and $t^{\mathrm{\prime }}$ to some common neighbourhood of $f(x)$. Thus there is a map $\mathrm{\Psi }:(f_{\mathrm{p}}\mathcal{G}{)}_x⟶{\mathcal{G}}_{f(x)}\mathrm{.}$

The maps $\mathrm{\Phi }$ and $\mathrm{\Psi }$ are inverse on representatives, so $(f_{\mathrm{p}}\mathcal{G}{)}_x\cong {\mathcal{G}}_{f(x)}$. Step 1.1 then gives $(f^{-1}\mathcal{G}{)}_x\cong {\mathcal{G}}_{f(x)}$.