Gaining insight into the Inverse Image Sheaf

Let $f: X \rightarrow Y$ be a continuous map of topological spaces and let $G$ be a sheaf of sets on $Y$. I am trying to understand the definition of the inverse image sheaf $f^{-1}G$ on $X$. This is the sheaf associated to the preshief $U \rightarrow lim_{V \supseteq f(U)} G(V)$ where $V$ is open in $Y$ and $U$ open in $X$. In particular i am trying to understand the quantity $lim_{V \supseteq f(U)} G(V)$. As i understand, this is a colimit in the category of sets. I read the category-theoretic definition of the colimit using co-cones, but i am having a hard time making a connection/interpretation. Also, is this a filtered colimit or just a colimit?

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It's a filtered colimit – if two open sets contain $f(U)$, then so does their intersection. –  Zhen Lin Sep 26 '12 at 1:36
@ZhenLin: Thanks, i am gonna work with this hint... –  Manos Sep 26 '12 at 2:15

1) First of all you have the crucial property $(f^{-1}G)_x=G_{f(x)}$ for all $x\in X$.
In particular, applying this to the inclusion $i:\lbrace x\rbrace \hookrightarrow X$, of a point, you get the interesting equality $(i^{-1}G)_x=G_x$ which shows that taking an inverse image is a generalization of taking the stalk of a sheaf at a point.

2) This is a situation where the pre-Grothendieck interpretation of a sheaf as an étalé space is quite illuminating:
If $Et(G)\to Y$ is the étalé space corresponding to $G$, take the fiber product $Z\stackrel {def}{=}X\times _Y Et(G)\to X$ in the category of topological spaces.
This is an étalé space over $X$ and the corresponding sheaf of sections is the sheaf on $X$ you are looking for: $Sh_Z=f^{-1}G$.

3) Beware that in the category of schemes (or in any other geometric category like that of analytic spaces) the important pull-back for sheaves of $\mathcal O_Y$-modules over $Y$ is not the topological one we are discussing but the algebraic one, namely $$f^{*}G =f^{-1}G \otimes _{f^{-1}(\mathcal O_Y)} \mathcal O_X$$
The simplest contrasting example is that of a scheme $X$ over a field $k$ and the inclusion of a rational point $i:\lbrace x\rbrace \hookrightarrow X$. There is a huge difference between both pull-backs of $\mathcal O_X$:
$$(i^{-1}\mathcal O_X)_x = \mathcal O_{X,x}\neq (i^{*}\mathcal O_X)_x =k$$

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+1 Georges. I only realised the power of passing to the stalk when trying to prove that sheafification commutes with restriction of a sheaf to an open set! –  Benja Jul 8 at 11:20