Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,O_X), (Y,O_Y)$ be ringed spaces where e.g. for any open set $U$ of $X$, $O_X(U)$ is a set of $k$-valued functions on $U$, called regular. Let $f:X \rightarrow Y$. I want to show that $f$ preserves regular functions under pullback.

Let $V$ be an open set of $Y$ and $\xi:Y \rightarrow k$ a regular function, i.e. $\xi \in O_Y(V)$. The pullback of $\xi$ under $f$ is $\xi \circ f|_{f^{-1}(V)}$. Now, $f^{-1}(V)$ is an open set of $X$, since $f$ is continuous. How do i proceed now? Why would $\xi \circ f|_{f^{-1}(V)} \in O_X(f^{-1}(V))?$ It seems that i have nothing to connect regular functions of $X$ with regular functions of $Y$.

share|improve this question
    
This is not how it works. In general, you can make sense of the idea of a "sheaf of functions" $O_X$, such that any continuous function $f: X \to Y$ automatically yields a (unique, natural) pullback $f^{\#}: O_Y \to f_* \mathcal{F}_X$, where $\mathcal{F}_X$ denotes the sheaf of discontinuous functions on $X$ (i.e. the sheaf such that $\mathcal{F}(U) = k^U$). There is also a natural inclusion $O_X \to \mathcal{F}$, but asking for $f^{\#}$ to take values in $O_X$ is (sometimes) a non-trivial additional condition, as can easily be seen by examples. –  Tom Bachmann Sep 24 '12 at 18:41
1  
For example, if $X, Y$ are smooth manifolds and $O_X, O_Y$ are the sheaves of smooth functions ($k = \mathbb{R}$), then $f^{\#}$ has this property iff $f$ is smooth in the classical sense. Similarly, if $k$ is an algebraically closed field, $X, Y$ are affine varieties and $O_X, O_Y$ the sheaves of regular functions, then $f^{\#}$ has this property iff $f$ is given by a polynomial. –  Tom Bachmann Sep 24 '12 at 18:42
    
@TomBachmann: Regarding your first comment, why is the pullback regular automatically? If $f$ is a morphism of the sheaves $O_X \rightarrow O_Y$, then by definition this is true. If $f$ however is only continuous, then how do we show the existence of this $f^{\#}$? This is precisely what i want to prove: that any continuous map $X \rightarrow Y$ is also a morphism of ringed spaces. –  Manos Sep 24 '12 at 18:50
1  
But it is not. I was just saying: if $\mathcal{F}_X$ denotes the sheaf such that $\mathcal{F}(U) = k^U$ (arbitrary functions from $U$ to $k$), then you seem to be working with "sheaves of functions", that is sheaves which come with a natural inclusion into $\mathcal{F}_X$. Clearly, if $f$ is continuous, then there exists the pullback $f^{\#}: \mathcal{F}_Y \to f_* \mathcal{F}_X$. Composing with the inclusion $\mathcal{O}_Y \to \mathcal{F}_Y$ yields $\mathcal{O}_Y \to f_* \mathcal{F}_X$, but this need not take values in $f_* \mathcal{O}_X$. –  Tom Bachmann Sep 24 '12 at 19:12
1  
One would typically define a "morphism of ringed spaces" to be a continuous map $f$ such that $f^\#$ does take values in $f_* \mathcal{O}_X$, but as I said before, this definition is not vacuous. [Moreover, in the typical definition of ringed spaces, we usually don't assume that $\mathcal{O}_X$ is given with an inclusion into $\mathcal{F}_X$, and so $f^\#$ must be given independently - although in certain cases it is necessarily of the form you describe.] –  Tom Bachmann Sep 24 '12 at 19:15
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.