inverse image presheaf property

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Given a continuous function $f:X\to Y$ of topological spaces, I'm trying to understand exactly what is the inverse image sheaf, I have read a particular property $(f^{-1}G)_x=G_{f(x)}$, where $G$ is a sheaf on $Y$. How can I get that identity on stalks if the left one is a stalk on a sheaf over $X$, and the right on is over $Y$?

I see this:

$(f^{-1}G)_x=\{[U,s], x\in U\subset X, s\in f^{-1}G(U)\}=\{[U,s], x\in U\subset X, s\in colim \{G(V), f(U)\subset V\}\}$

and

$(G)_{f(x)}=\{[V,s], f(x)\in V\subset Y, s\in G(V)\}$

So how can this two of this classes be equivalent if they are on different topological spaces? or it is only an isomorphism?.

Sorry for the "colimit" word, I look for a command but did not found one.

Appreciate any help.

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$(f^{-1}G)_x= \lim_{U \ni x} (f^{-1}G)(U)=\lim_{U \ni x}( \lim_{V \supseteq f(U)} G(V))$.

Now you can identify $\lim_{U \ni x}( \lim_{V \supseteq f(U)} G(V))$ with $(G)_{f(x)}$, as every $V$ such that $ V \supseteq f(U)$ contains $f(x)$ obviously: Elements in $\lim_{U \ni x}( \lim_{V \supseteq f(U)} G(V))$ could be seen as $[U',[V',s]]$, where $U'\ni x$, $V' \supseteq f(U')$ and $s\in G(V')$. You would map such an element to $[V',s]\in (G)_{f(x)} $, as $V'\ni f(x)$. This is well-defined and gives you $(f^{-1}G)_x=G_{f(x)}$.