Open vs Closed Immersions of Locally Ringed Spaces I'm reading Qing Liu's book at the moment and I'm trying to figure out why open immersions of locally ringed spaces are required to be isomorphisms on stalks, but closed immersions are only required to be surjective on stalks. I searched the site, googled a bit, and doodled around with the nature of sheaves not being "nice" with respect to images (i.e. needing sheafification in general), but no luck. Any good explanations?
 A: Thanks to Hoot's remark, I tried working it out in more specific case. Let $k$ be a field, and define $A=k[x,y]$, $f=x^2+y^2-1$, $B=A_f$, $C=A/(f)$, $X=\mathbb{A}_k^2$, $Y=X_f\cong Spec(A_f)$, and $Z=V(f)\cong Spec(A/(f))$. 
Now I'll let $\phi:Y\rightarrow X$ and $\psi:Z\rightarrow X$ be the maps induced from the natural maps $A\rightarrow B$ and $A\rightarrow C$. I already know that $\phi$ and $\psi$ are homeomorphisms onto their images, the image of $\phi$ is open in $X$, and the image of $\psi$ is closed. Furthermore, these are just the inclusion of the unit circle and its complement into the plane, so if anything should be an open/closed immersion, these should.
If I let $g\in A$, then $\mathcal{O}_X(X_g)=A_g$, $$(\phi_*\mathcal{O}_Y)(X_g)=\mathcal{O}_Y(\phi^{-1}(X_g))=\mathcal{O}_Y(Y_{\phi(g)})=A_{fg},$$ and $$(\psi_*\mathcal{O}_Z)(X_g)=\mathcal{O}_Z(\psi^{-1}(X_g))=\mathcal{O}_Z(Z_{\psi(g)})=(A/(f))_g.$$
Now let $\mathfrak{p}$ be a prime of $A$ not containing $fg$. Since $\phi(Y_g)=X_{fg}=X_f\cap X_g\subseteq X_g$, and $g$ was an arbitrary element of $A$ not in $\mathfrak{p}$ (I know you could complain at the order I introduced the variables, but it is immaterial here), the direct limit of $A_{fg}$ over all $g$ is actually $A_{\mathfrak{p}}$. So the map $\phi^{\#}_{\mathfrak{p}}$ on stalks at $\mathfrak{p}$ is actually just the identity map on $A_{\mathfrak{p}}$. So, if I assume that this case is somewhat generic, I should expect an open immersion to be an isomorphism on stalks.
On the other hand, let $\mathfrak{q}$ be a prime of $A$ containing $f$ but not $g$. Passing to stalks, $\psi^{\#}_{\mathfrak{q}}$ is the quotient map $A_{\mathfrak{q}}\rightarrow A_{\mathfrak{q}}/(f)_{\mathfrak{q}}$. So $\psi^{\#}$ is surjective on stalks, but won't be an isomorphism unless $f=0$ (since $A$ is a domain).
Now the issue is moral. All I can vaguely sense is that an open immersion preserves dimension, whereas a closed immersion only preserves dimension (in this case) when it is the identity map. Maybe this is because open sets are dense in $\mathbb{A}_k^2$, so a dimension preserving closed immersion should be the whole space? On the other hand, if a generic closed immersion hits some subvariety of the space (as in this case), then the restriction to the subvariety reduces the number of distinct sections at a point. So the map on germs will not be injective. Anyway, thanks for the suggestion. I know it's simple, but I haven't been doing this topic long enough to have good habits yet. Also, I would appreciate any other remarks on whether or not this is the right idea.
