It is usually known the fact that the pullback sheaf preserves stalks in the following way: if $f:X\to Y$ is a continuous map of topological spaces and $\mathcal{F}$ is a sheaf of sets over $Y$, then $(f^{-1}\mathcal{F})_x=\mathcal{F}_{f(x)}$. What I haven't seen mentioned on the literature is if $f^{-1}$ also preserves induced map on stalks. Specifically, does the diagram
commute? (up to isomorphism of functors, say) What I stated is commutativity on objects. But why on morphisms? For a sheaf of sets $\mathcal{F}$ over $Y$, denote $f_p\mathcal{F}$ to the sheaf of sets over $X$ from which $f^{-1}\mathcal{F}$ comes (see for example 008F for the precise definition of $f_p$). Consider the diagram
Where $(-)^\#$ is the sheafification functor. The upper triangle of the diagram commutes (by definition of $f^{-1}$). Since the composite $$ \operatorname{PSh}_\mathsf{Set}(X)\xrightarrow{(-)^\#}\operatorname{Sh}_\mathsf{Set}(X)\xrightarrow{(-)_x}\mathsf{Set} $$ equals $(-)_x:\operatorname{PSh}_\mathsf{Set}(X)\to\mathsf{Set}$, to see commutativity of the lower triangle in last diagram, it suffices to see commutativity of
But now, why does this diagram commutes? I never fully understood well the proof of $(f^{-1}\mathcal{F})_x=\mathcal{F}_{f(x)}$. I think the proof uses the fact that colimits commute with colimits, but I think also something else. I don't know how to upgrade the proof to the case of induced maps on stalks.