The identification $(\pi^{-1}\mathcal{G})_p \to \mathcal{G}_{\pi(p)} $ induces a continuous map between the "space of sections" Let $\pi:X\to Y$ be a continuous map and $\mathcal{G}$ be a sheaf on $Y$, then there is a natural isomorphism $f_p:(\pi^{-1}\mathcal{G})_p \to \mathcal{G}_{\pi(p)} $ (by using the adjunction of inverse image and pushforward or whatever means). $f_p's$ will induce a map $f: \bigsqcup_{p\in X}(\pi^{-1}\mathcal{G})_p \to \bigsqcup_{q\in Y}\mathcal{G}_{q}$.
I wonder if this map $f$ is also a continuous map with respect to the "space of section" (espace etale) topology, whose base are sets $\{(x,s_x):x\in U\}$, where $U\subset X$ is open and $s\in \mathcal G(U)$. The main difficulty for me is to find the explicit formula for $f_p$'s.
I ask this because I need this result to prove: 
The pullback of the "space of sections" is the same as the "space of sections" of the pullback

Yes it is. Moreover, if we let $\mathcal E=\coprod_{q\in Y}\mathcal{G}_q$ be the etale space of $\mathcal G$ and $\mathcal{E}'
=\coprod_{p\in X}\mathcal{G}_{\pi (p)}$ be the etale space of $\pi^{-1}\mathcal G$
then we have a commutative square
$\require{AMScd}$
\begin{CD}
\mathcal{E'} @>f>> \mathcal{E}\\
@V  V V @VV  V\\
X @>>\pi> Y
\end{CD}
which is a pullback (fibre product) in the category of topological spaces. Indeed we can define the inverse image of sheaf maps by insisting
that $\mathcal{E}'=\mathcal{E}\times_Y X$ be the fibre product.
