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Let $X$ and $Y$ be Riemann surfaces, and $\mathscr{O}_X, \mathscr{O}_Y$ be the sheaves of holomorphic functions on $X$ and $Y$ respectively.

It is obvious that a holomorphic map $f:X \to Y$ gives rise to a morphism of locally ringed spaces $(X, \mathscr{O}_X) \to (Y, \mathscr{O}_Y)$.

I was always under the impression that a morphism of locally ringed spaces also gave rise to a holomorphic map, but I can't for the life of me prove it. I cant see why the induced stalk maps being local homomorphisms forces the topological map to be holomorphic.

Any help would be appreciated!

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up vote 5 down vote accepted

If I am not mistaken, for this you need to consider $(X,\mathcal O_X)$ and $(Y,\mathcal O_Y)$ not as "naked" locally ringed spaces, but as spaces locally ringed by $\mathbb C$-algebras, and thus consider morphisms $(X,\mathcal O_X) \to (Y,\mathcal O_Y)$ that respect the $\mathbb C$-algebra structure on the sheaves of rings.

With this assumption, it is not hard to show that for such a morphism $(f,f^{\natural})$ (so $f:X \to Y$ is continuous, and $f^{\natural}: \mathcal O_Y \to f_* \mathcal O_X$ is a map of sheaves of $\mathbb C$-algebras which induces local maps on stalks), the map $f^{\natural}$ on sheaves is necessarily given by pull-back of functions.

Thus $f:X \to Y$ is a continuous map which pulls back holomorphic functions on open subsets $U$ of $Y$ to holomorphic functions on the preimage of $U$ in $X$. Applying this to the holomorphic coordinate functions on coordinate charts of $Y$, one finds that $f$ is in fact a holomorphic map, as required.

(Note: there is nothing special about dimension one here; this will work for complex manifolds of any dimension.)

(Note also that Soarer gives the same answer more succinctly!)

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The part that i was struggling with was showing that the sheaf map was necessarily given by pull back. Once you know that it is clear. - edit: I just figured it out. thanks for the help though! – DBr Jan 13 '11 at 6:14

take a chart of $Y$, and let $y$ be the coordinate function. First show that the morphism actually comes from a honest map from $f: X \to Y$, and that the sheaf map takes $y$ to $y \circ f$. Now $y \circ f$ lying in the sheaf of holomorphic functions on $X$ means exactly that $f$ is holomorphic locally. Varying the chart shows that $f$ is holomorphic.

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What I was actually stuck on was showing that the sheaf map is given by pull back. It is painfully simple...

You need to assume that $f^\sharp $ is a morphism of sheaves of $\mathbb{C}$-algebras as pointed out by Matt E

Assume $(f,f^\sharp ):X \to Y$ is a morphism of locally ringed spaces.

Fix $s \in \mathscr{O}_Y(U)$. Fix $p \in U$. Then $s_{f(p)} - (s \circ f(p))_{f(p)}$ is in the maximal ideal of $\mathscr{O}_{Y,f(p)}$. Therefore $f^\sharp (U)(s)_p - (s \circ f(p))_p$ is in the maximal ideal of $\mathscr{O}_{X,p}$. Therefore $f^\sharp (U)(s)(p) = s \circ f(p)$

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