Topological Ring and Ringed Space

IntroductionFunctor to RingSheaf and Ringed Space

 

Introduction

In mathematical analysis, given $f,g\in C(\mathbb{R})$$f,g\in C(\mathbb{R})$, we need to prove that $f+g,f\cdot g\in C(\mathbb{R})$$f+g,f\cdot g\in C(\mathbb{R})$ as well. But using epsilon-delta language means using some trick, and I do not want to use any trick. Let us generalize this question to ${\mathrm{Hom}}_{\mathrm{Top}}(X,R)$$\mathrm{Hom}_{\mathrm{Top}}(X,R)$ for arbitrary topological space $X$$X$ and topological ring $R$$R$. Also, restricting the domain of the hom functor to $\mathrm{Op}(X{)}^{\mathrm{op}}$$\mathrm{Op}(X)^{\mathrm{op}}$ will give us a sheaf of continuous functions over $X$$X$, denoted as ${\mathcal{O}}_X$$\mathcal{O}_X$, and a ringed space $(X,{\mathcal{O}}_X)$$(X,\mathcal{O}_X)$​​.

Functor to Ring

Lemma. Let $Y,Z$$Y,Z$ be two topological spaces. Then the constant function $\mathrm{\forall }y\in Y,f(y)=z$$\forall y\in Y, f(y)=z$ is continuous.

Proof. Let $V\subseteq Z$$V\subseteq Z$ be an open set, then either $z\in V$$z\in V$ or $z\notin V$$z\notin V$. If $z\in V$$z\in V$ then $f^{-1}(V)=X$$f^{-1}(V)=X$. If $z\notin V$$z\notin V$ then $f^{-1}(V)=\mathrm{\varnothing }$$f^{-1}(V)=\emptyset$. $◻$$\square$

Let $X$$X$ be an arbitrary topological space and $(R,+,\cdot )$$(R,+,\cdot)$ a topological ring (Ring object in $\mathrm{Top}$$\mathrm{Top}$).

Proposition. ${\mathrm{Hom}}_{\mathrm{Top}}(-,R)$$\mathrm{Hom_{Top}}(-,R)$ gives us a functor from $\mathrm{Top}$$\mathrm{Top}$ to $\mathrm{Ring}$$\mathrm{Ring}$.

Proof. We need to prove that $0,1\in {\mathrm{Hom}}_{\mathrm{Top}}(X,R)$$0,1\in \mathrm{Hom_{Top}}(X,R)$, and for $f,g\in {\mathrm{Hom}}_{\mathrm{Top}}(X,R)$$f,g\in \mathrm{Hom_{Top}}(X,R)$, $f+g,f\cdot g\in {\mathrm{Hom}}_{\mathrm{Top}}(X,R)$$f+g,f\cdot g\in \mathrm{Hom_{Top}}(X,R)$.

It is easy to see that $0,1\in {\mathrm{Hom}}_{\mathrm{Top}}(X,R)$$0,1\in \mathrm{Hom_{Top}}(X,R)$ since these are constant functions.

The universal property of the product $R\times R$$R\times R$ claims that $(f,g):X\to R\times R$$(f,g):X\to R\times R$ is continuous, and by the definition of a topological ring, $+,\cdot :R\times R\to R$$+,\cdot: R\times R\to R$ are continuous as well.

Therefore, $f+g,f\cdot g$$f+g,f\cdot g$ are continuous since they are compositions of continuous functions. $◻$$\square$

Now the Hom functor gives us a presheaf ${\mathrm{Op}}^{\mathrm{op}}(X)\to \mathrm{Ring}$$\mathrm{Op^{op}}(X)\to\mathrm{Ring}$. The restriction maps are given by pullback $i^{\ast }(f)=f\circ i$$i^*(f)=f\circ i$.

We denote ${\mathrm{Hom}}_{\mathrm{Top}}(U,R)$$\mathrm{Hom}_{\mathrm{Top}}(U,R)$ as $F(U)$$F(U)$​.

Sheaf and Ringed Space

Proposition. The presheaf given by the Hom functor is a sheaf.

Proof.

Let $U\subseteq X$$U\subseteq X$ be an open set and $(U_i{)}_{i\in I}$$(U_i)_{i\in I}$ be an open covering of $U$$U$.

Let $(f_i{)}_{i\in I}$$(f_i)_{i\in I}$ be a family of functions, with $f_i\in F(U_i)$$f_i\in F(U_i)$, satisfying $f_i(U_i\cap U_j)=f_j(U_i\cap U_j)$$f_i(U_i\cap U_j)=f_j(U_i\cap U_j)$. Then there exists a unique function $f\in F(U)$$f\in F(U)$ such that ${\mathrm{res}}_{U_i}^U(f)=f_i$$\mathrm{res}_{U_i}^U(f)=f_i$.

This follows directly from the fact that $(U_i{)}_{i\in I}$$(U_i)_{i\in I}$ is an open covering of $U$$U$. To see that $f$$f$ is continuous, let $V$$V$ be an open set in $R$$R$ and consider $f^{-1}(V)$$f^{-1}(V)$:

$$\begin{array}{}\text{(1)}& f^{-1}(V)=\{x\in X:f(x)\in V\}=\bigcup _{i\in I}\{x\in U_i:f(x)\in V\}=\bigcup _{i\in I}\{x\in U_i:f_i(x)\in V\}=\bigcup _{i\in I}{f}_i^{-1}(V)\end{array}$$

Hence, we get a ringed space $(X,{\mathcal{O}}_X)$$(X,\mathcal{O}_X)$.