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$\newcommand{\Spec}{\operatorname{Spec}} \newcommand{\O}{\mathscr{O}}$ Let $X=\Spec A$, and $Y=\Spec B$, and suppose that $f:X\rightarrow Y$ is a morphism coming from the ring homomorphism $\phi:B\rightarrow A$. This induces sheaf morphism $f^\sharp:\O_Y\rightarrow (f_*\O_X)$ which is given by localizations of $\phi$ at elements in $b\in B$, and $\phi(b)\in A$. It is easy to check that the stalk map $(\O_Y)_y\rightarrow (f_*\O_X)_y$, is then given by the induced map $B_\mathfrak p\rightarrow A_\mathfrak p$, where we are taking $A$ to be a $B$ module, and $\mathfrak p$ is the prime ideal corresponding to the point $y$. However, what about the stalk map $(\O_Y)_{f(x)}\rightarrow (\O_X)_x$?

In particular, for arbitrary locally ringed spaces, let $f:X\rightarrow Y$ is a morphism of local rings, and $f_{f(x)}^\sharp:(\O_Y)_{f(x)}\rightarrow (f_*\O_X)_{f(x)}$ be the unique stalk map given on equivalence classes $[U,s]_{f(x)}\in (\O_Y)_{f(x)}$ by $[U,s]_{f(x)}\rightarrow [U,f^\sharp_U(s)]_{f(x)}$. Then I would define $f_x:(\O_Y)_{f(x)}\rightarrow (\O_X)_x$ as the composition of the map of the above map with the map $(f_*)_x:[U,s]_{f(x)}\in (f_*\O_X)_{f(x)}\mapsto [f^{-1}(U),s]_x\in (\O_X)_x$. In other words all that last map is changing is how restrictive our equivalence relation is.

What is the analogue of this in the case of affine schemes? If $\mathfrak p=\phi^{-1}(\mathfrak q)$ is there somehow a natural morphism $A_\mathfrak p\rightarrow A_\mathfrak q$ that would clearly agree with the above construction?

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The morphism $B_{\mathfrak{p}} \to A_{\mathfrak q}$ that you're looking for comes from the composition $B \to A \to A_{\mathfrak{q}}$ and the universal property of localization. All you need to check to get an induced morphism is that if $b \in B \setminus \mathfrak{p}$, then $b$ gets mapped to a unit in $A_{\mathfrak{q}}$. But if $b \in B \setminus \phi^{-1}(\mathfrak{q})$, then $\phi(b) \notin \mathfrak{q}$, so $\frac{\phi(b)}{1}$ is a unit in $A_{\mathfrak{q}}$

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  • $\begingroup$ How do we see that this agrees with the abstract version of this morphism? $\endgroup$
    – Chris
    Commented May 22 at 14:29
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In total generality if you have spaces $X,Y$ and a continuous map $f:X\to Y$ and a given morphism of sheaves $\phi:\mathscr{F}\to f_\ast\mathscr{G}$ for some sheaves $\mathscr{F},\mathscr{G}$ on $Y,X$ respectively, we can get our hands on maps $\mathscr{F}_{f(x)}\to\mathscr{G}_x$ using formal nonsense only. I think it is instructive to see this; it makes the construction feel less improvised.

Firstly, there is a map $\phi_{f(x)}:\mathscr{F}_{f(x)}\to(f_\ast\mathscr{G})_{f(x)}$ and it remains to find some kind of canonical homomorphism $(f_\ast\mathscr{G})_{f(x)}\to\mathscr{G}_x$. But indeed, $(f_\ast\mathscr{G})_{f(x)}\cong(f^\ast(f_\ast\mathscr{G}))_x$ and there is a canonical counit $f^\ast f_\ast\mathscr{G}\to\mathscr{G}$ which restricts to a map of the form we want.

Less abstractly, if you think about this isomorphism and how the counit is actually defined, this homomorphism works as you say. If $\xi\in(f_\ast\mathscr{G})_{f(x)}$ then it is represented as the germ of some $\alpha\in(f_\ast\mathscr{G})(V)=\mathscr{G}(f^{-1}V)$ where $V$ is an open neighbourhood of $f(x)$ and thus $f^{-1}V$ an open neighbourhood of $x$; the germ of $\alpha$ in $Y$, $\xi$, is mapped to the germ of $\alpha$ in $X$ i.e. viewing it as a section of $\mathscr{G}$. Though it feels like this does nothing, this homomorphism need not in general inject or surject I don't think. It does under some point-set assumptions though; what you really need is that the neighbourhoods of form $f^{-1}V$ are cofinal and this happens if the spaces involved are LCH and $f$ is (quasi)proper - not a situation common to algebraic geometry.

If $X,Y$ are affine schemes, spectra of $A,B$ respectively and $\mathscr{F},\mathscr{G}$ their structure sheaves and $(f,\phi)$ a scheme morphism we know that, really, $\phi$ is induced by a homomorphism $g:B\to A$ and for a given point $\mathfrak{p}$ of $A$, we need to understand the map $(f_\ast\mathscr{O}(X))_{g^{-1}\mathfrak{p}}\to\mathscr{O}(X)_\mathfrak{p}\cong A_\mathfrak{p}$. The general element of the left hand side is the germ of some section of $\mathscr{O}(X)(f^{-1}D(u))=\mathscr{O}(X)(D(g(u)))$, where $g^{-1}\mathfrak{p}\in D(u)$ i.e. $g(u)\notin\mathfrak{p}$, and therefore is the germ of some $a/g(u)$ (replacing $u$ with some power of $u$ if necessary) where $a\in A$ but be mindful this is a germ at a point of $Y$, not $X$. This element $[a/g(u)]$ is identified (via the isomorphism and canonical counit) with, well, "the same thing" i.e. the germ of $a/g(u)$ at $\mathfrak{p}$. So really, the overall induced map $\mathscr{O}(Y)_{f(\mathfrak{p})}\cong B_{g^{-1}\mathfrak{p}}\to A_\mathfrak{p}\cong\mathscr{O}(X)_{\mathfrak{p}}$ on stalks is just $[b/u]\mapsto[g(b)/g(u)]_{\text{ in $Y$}}\mapsto[g(b)/g(u)]_{\text{ in $X$}}$. The first map is just by definition of $\phi$ the sheaf map associated to the ring map $g$. Leaving equivalence classes understood, we just write this as $b/u\mapsto g(b)/g(u)$.

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