Let $f:X\to Y$ be a continuous map, and let $\mathscr G$ be a sheaf on $Y$. There is a canonical morphism $\varphi:\mathscr G\to f_*f^{-1}\mathscr G$, hence a map $\varphi_y:\mathscr G_y\to f_*f^{-1}\mathscr G_y$, for any $y\in Y$. If $x\in X$, is it true that $\varphi_{f(x)}$ is an isomorphism?
I explain why I think it's true. By definition $\mathscr G_{f(x)}=\operatorname{colim}_{U\ni f(x)}\mathscr G(U)$; $f_*f^{-1}\mathscr G_{f(x)}$
instead is $$\operatorname{colim}_{U\ni f(x)}\operatorname{colim}_{V\supset ff^{-1}(U)}\mathscr G(V) .$$ Given an open $f(x)\in U\subset Y$, any open $ff^{-1}(U)\subset V\subset Y$ contains $f(x)$; moreover $ff^{-1}(U)\subset U$, so the double colimit is (isomorphic to) the colimit over the system of open sets containing $f(x)$, i.e. $\mathscr G_{f(x)}$. Since the map $\varphi_{f(x)}$ is compatible with the legs of the colimits, it is a (unique) isomorphism.
My doubt arises from an exercise (2.13) in Liu's algebraic geometry book; he asks to prove that, in the same hypothesis as above, $\varphi$ is surjective if $f$ is a closed immersion. If the image of $f$ is closed (but $f$ is not necessarily an immersion),
$\varphi_y$, for $y\notin f(X)$, is surjective as its codomain is the terminal object; so from the argument of the paragraph above, it seems that $\varphi$ is surjective just if the image of $f$ is closed. Likely, the result of the paragraph above needs the assumption of $f$ being an immersion, but I don't understand why. Thank you
Edit. Let $f:X\to Y$ be a continuous map. I'll denote the sheafication of a presheaf $\mathscr P$ with $\mathscr P^\#$, the pullback of a sheaf $\mathscr G$ on $Y$ with $f_p\mathscr G$, and the sheafication of the pullback with $f^{-1}\mathscr G$; I'll denote by $\varphi$ the canonical morphism $\mathscr G\to f_*f^{-1}\mathscr G$. Above I proved that $$\mathscr G_{f(x)}\to (f_*f_p\mathscr G)_{f(x)}, $$ and thus $$\mathscr G_{f(x)}\to (f_*f_p\mathscr G)_{f(x)}\to (f_*f_p\mathscr G)_{f(x)}^\#, \ (*)$$ are isomorphisms for all $x\in X$. Instead I have to consider $$\varphi_{f(x)}:\mathscr G_{f (x)}\to (f_*f^{-1}\mathscr G)_{f(x)},$$ different from $(*)$ because, as you say, pushforward doesn't commute with sheafication. If $f$ is an immersion, $(f_*\mathscr F)_{f(x)}\to \mathscr F_x$ is an isomorphism for any sheaf $\mathscr F$ on $X$ and any $x\in X$. Also, $\varphi_{f(x)}$ factors as $$\mathscr G_{f(x)}\to (f^{-1}\mathscr G)_x\to (f_*f^{-1}\mathscr G)_{f(x)},$$ which is a composition of isomorphisms. This shows that $\varphi_{f(x)}$ is an isomorphism, and if $f(X)$ is closed, $\varphi_y$ is trivially surjective for all $y\notin f(X)$, guaranteeing that $\varphi_y$ is surjective for all $y\in Y$.
Best Answer
I'll expand on Zhen Lin's comment since it's a nice simple example and you seem to be making the same mistake when looking at that example as you are in the main part of the question.
I think the core problem here is that you're forgetting that you need to sheafify the colimit when defining the inverse image functor.
I.e. $f^{-1}\mathscr{G}$ is a sheaf on $X$ defined as the sheafification of the presheaf $$\operatorname{colim}_{V\supseteq f(U)} \mathscr{G}(V).$$
It's also worth pointing out that sheafification doesn't commute with pushforward in general, so you can't compute the stalk of the pushforward of the inverse image with a colimit in the way that you want. After all, if you could commute the pushforward with the sheafification, then when taking the stalks you could drop the sheafification, and you'd be left with the colimit in your question, so this wouldn't be an issue. But they don't commute, so the argument in your question doesn't apply.
Now let's look at Zhen Lin's suggestion from the comments, where we take $X$ discrete and $Y$ a point.
A sheaf on $Y$ is defined by just the set $\mathscr{G}(Y)$. Similarly since $X=\{x_1,\ldots,x_n\}$ is discrete (and I'm going to just let it be finite for convenience's sake), a sheaf on $X$ is defined by the sets $\mathscr{H}(\{x_i\})$ for $1\le i \le n$. In particular, the sheafification of a presheaf $\mathscr{H}$ is easy to compute. It's just $$\mathscr{H}^\#(\{x_{i_1},\ldots,x_{i_k}\})=\prod_{j=1}^k \mathscr{H}(\{x_{i_j}\})$$
So to compute $f^{-1}\mathscr{G}$, we just need to compute $$f^{-1}\mathscr{G}(\{x_i\}) = \operatorname{colim}_{V: x_i\in f^{-1}(V)}\mathscr{G}(V)$$ Since $Y$ is a point, the only nonempty open set is $Y$ itself, so we have $$f^{-1}\mathscr{G}(\{x_i\}) = \mathscr{G}(Y).$$ (Note that we can also see this from the fact that these sets are also the stalks at the points, since it's always true that the stalks of the inverse image at any point $x$ are isomorphic to the stalks of the original sheaf at $f(x)$.)
Then $f_*f^{-1}\mathscr{G}$ is determined by its value on $Y$, which is $f^{-1}\mathscr{G}(X)$, which is $\mathscr{G}(Y)^n.$ Note that this is also the stalk at the unique point in $Y$, so the natural map cannot be an isomorphism on stalks.