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Let $(X, \mathcal O_X), (Y, \mathcal O_Y)$ be locally ringed spaces. A morphism of ringed spaces is defined to be a pair $(f,f^{\#}):(X, \mathcal O_X) \rightarrow (Y, \mathcal O_Y)$, where $f:X \rightarrow Y$ is continuous, and $f^{\#}: \mathcal O_Y \rightarrow f_{\ast} \mathcal O_X$ is a morphism of sheaves. We consider $(f,f^{\#})$ to also be a morphism of locally ringed spaces if for each $x \in X$, the homomorphism on the stalks $$f_x^{\#}: \mathcal O_{Y,f(x)} \rightarrow \mathcal O_{X,x}$$ is a local homomorphism (preimage of the unique maximal ideal remains maximal). My question is, what exactly is the map $f_{x}^{\#}$? I know since $f^{\#}$ is a morphism of sheaves, we have a homomorphism on the stalks $$f_{f(x)}^{\#}: \mathcal O_{Y,f(x)} \rightarrow (f_{\ast} \mathcal O_X)_{f(x)}$$ Are we getting $f_x^{\#}$ by composing $f_{f(x)}^{\#}$ with some homomorphism $(f_{\ast} \mathcal O_X)_{f(x)} \rightarrow \mathcal O_{X,x}$?

Eric Wofsey
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D_S
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    You really mean to write $(f_*\mathcal{O}X){f(x)}$. Then there is a map as desired. It should be the only thing you can write down. – Hoot Dec 11 '15 at 02:47

2 Answers2

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Given any sheaf $\mathcal{O}$ on a space $X$, a continuous map $f:X\to Y$, and a point $x\in X$, there is a canonical map $(f_*\mathcal{O})_{f(x)}\to\mathcal{O}_x$. Indeed, an element of $(f_*\mathcal{O})_{f(x)}$ is represented by a section of $f_*\mathcal{O}$ over some open set $V$ containing $f(x)$, which is just a section of $\mathcal{O}$ over $f^{-1}(V)$, which then determines an element of $\mathcal{O}_x$. It is easy to see that this correspondence is compatible with restriction and hence induces a well-defined map $(f_*\mathcal{O})_{f(x)}\to\mathcal{O}_x$.

In your case, taking $\mathcal{O}=\mathcal{O}_X$, this is the map you are looking for.

Eric Wofsey
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  • So one way to describe it, is for each open $V$ in $Y$ with the property that $x \in f^{-1}V$, we have a homomorphism $f_{\ast} \mathcal O_X (V) = \mathcal O_X(f^{-1}V) \rightarrow \mathcal O_{X,x}$, and hence a homomorphism from the direct limit $(f_{\ast} \mathcal O_X){f(x)} \rightarrow \mathcal O{X,x}$. – D_S Dec 11 '15 at 02:52
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There is another way to describe $f^{\#}_x: \mathcal{O}_{Y,f(x)}\rightarrow \mathcal{O}_{X,x}$

Let $(X, \mathcal{O}_X)$ and $(Y,\mathcal {O}_Y)$ be two schemes.

A map between between $(X, \mathcal{O}_X)$ and $(Y,\mathcal {O}_Y)$ is a pair

1)$f:X\rightarrow Y$ and

2)$f^{\#}:\mathcal{O}_Y\rightarrow f_{*}\mathcal{O}_X$

such that the induced map $f_x^{\#}: \mathcal O_{Y,f(x)} \rightarrow \mathcal O_{X,x}$ is a local homomorphism of rings.

Let us try to write down the induced map-

Note, $\mathcal{O}_{Y,f(x)}=\varinjlim_{ f(x)\in V}\mathcal{O}_Y(V)$

Now by 2) we have map $f^{\#} (V):\mathcal{O}_Y(V)\rightarrow f_{*}\mathcal{O}_X(V)=\mathcal {O}_X(f^{-1}(V))$

Now, as $f(x)\in V \implies x\in f^{-1}(V)$. Therefore, there exist maps $g_x(f^{-1}(V)):\mathcal {O}_X(f^{-1}(V))\rightarrow \mathcal{O}_{X,x}$ (property of direct limit)

Therefore the composition of this two maps

$\mathcal{O}_Y(V)\xrightarrow{f^{\#} (V)}\mathcal {O}_X(f^{-1}(V))\xrightarrow{g_x(f^{-1}(V))} \mathcal{O}_{X.x}$

Now, by the universal property of direct limit of $\mathcal{O}_{Y,f(x)}$ one gets an induced map from $\mathcal{O}_{Y,f(x)}\rightarrow \mathcal{O}_{X,x}$ which is the desired map $f^{\#}_x$ which we need to be local ring homomoorphsim

Universal Property of direct limit cum definition: Let $\{M_\lambda, f_{\lambda \mu}\}_{\lambda \in I}$ where $f_{\lambda \mu}:M_\lambda \rightarrow M_\mu$ for all $\lambda \leq \mu$ be a directed system of rings. A ring $M$ is said to be the directed limit of the directed system if

  1. There exists maps $g_{\lambda}:M_\lambda \rightarrow M$ for all $\lambda\in I$ such that $g_\lambda=g_\mu \circ f_{\lambda \mu}$ for all $\lambda\leq \mu$

  2. If there exists another ring $M'$ with maps $g'_{\lambda}:M_\lambda \rightarrow M'$ for all $\lambda\in I$ such that $g'_\lambda=g'_\mu \circ f_{\lambda \mu}$ for all $\lambda\leq \mu$

Then there exists a unique map from $M\rightarrow M'$

Babai
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