5
$\begingroup$

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 $(f_{\ast} \mathcal O_X)_{f(x)} \rightarrow \mathcal O_{X,x}$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Hoot Dec 11 '15 at 2:47
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $(f_{\ast} \mathcal O_X)_{f(x)} \rightarrow \mathcal O_{X,x}$. $\endgroup$ – D_S Dec 11 '15 at 2:52
2
$\begingroup$

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 $M\rightarrow M'$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.