A contravariant functor from subcat of Top to Algebra, a kind of algebra-geometry duality (Typo, in the lattice isomorphism, S should be the closed set, so we do not have the Boolean Homomorphism)

We already discuss some algebraic property of continuous function.

Math Essays: Continous function, a interesting view (wuyulanliulongblog.blogspot.com)

Now we will dive to the deep place.

Algebra

Consider the ring $C(\mathbb{R})$ $C(\mathbb R)$ . It is not Noetherian since $I_n:=\{f\in C[\mathbb{R}]\mid \mathrm{\forall }x\ge n,f(x)=0\}$ $I_n := \{f \in C[\mathbb R] \mid \forall x \ge n, f(x) = 0\}$ is not stable.

If we consider $C([a,b])$ $C([a,b])$ , it is also not Noetherian. Take, for example, $C([0,1])$ $C({[0,1]})$ and the sequence of intervals $[0,\frac{1}{n}]$ $[0,\frac{1}{n}]$ .

Now, you might wonder why if $C([0,1])$ $C({[0,1]})$ is not Noetherian, then $C([a,b])$ $C({[a,b]})$ is not either?

The answer lies in the fact that $[0,1]\cong [a,b]$ $[0,1] \cong [a,b]$ in the category of topology. In other words, they are homeomorphic.

As you can observe, homeomorphisms induce ring isomorphisms. Additionally, every subset of $[a,b]$ $[a,b]$ corresponds to an ideal. This suggests the existence of a functor $F:{\mathrm{Top}}^{\mathrm{op}}\to \mathrm{Ring}$ $F: \mathrm{Top}^{\mathrm{op}} \to \mathrm{Ring}$ . (What is the adjoint of $F$ $F$ ?)

Now, let's get back to the concrete matter at hand. Consider a continuous function $h:[a,b]\to [c,d]$ $h: [a,b] \to [c,d]$ .

We can define $h^{\ast }(g):=g\circ h$ $h^*(g) := g \circ h$ , which is a ring homomorphism from $C[c,d]$ $C{[c,d]}$ to $C[a,b]$ $C{[a,b]}$ because:

• $h^{\ast }(f+g)=(f+g)\circ h=f\circ h+g\circ h=h^{\ast }(f)+h^{\ast }(g)$ $h^*(f + g) = (f + g) \circ h = f \circ h + g \circ h = h^*(f) + h^*(g)$

• $h^{\ast }(fg)=(fg)\circ h=f(h)g(h)=h^{\ast }(f)h^{\ast }(g)$ $h^*(fg) = (fg) \circ h = f(h)g(h)=h^*(f)h^*(g)$

• $h^{\ast }(\lambda g)=(\lambda g)\circ h=\lambda (g\circ h)=\lambda h^{\ast }(g)$ $h^*(\lambda g) = (\lambda g) \circ h = \lambda(g \circ h) = \lambda h^*(g)$

Since $h$ $h$ is a homeomorphism, $h^{\ast }$ $h^*$ is a ring isomorphism. To prove this, let $g$ $g$ be the inverse of $h$ $h$ , such that $h\circ g={\mathrm{id}}_{C[c,d]}$ $h \circ g = \mathrm{id}_{C{[c,d]}}$ and $g\circ h={\mathrm{id}}_{C[a,b]}$ $g \circ h = \mathrm{id}_{C{[a,b]}}$ . Then, $g^{\ast }\circ h^{\ast }={\mathrm{id}}_{C[c,d]}$ $g^* \circ h^* = \mathrm{id}_{{C{[c,d]}}}$ and $h^{\ast }\circ g^{\ast }={\mathrm{id}}_{C[a,b]}$ $h^* \circ g^* = \mathrm{id}_{C{[a,b]}}$ .

Actually we can consider the functor $\mathrm{Hom}(-,\mathbb{R})$ $\mathrm{Hom}(-,\mathbb R)$ , it maps topology space $X$ $X$ to Algebra $C(X)$ $C(X)$ , and continuous function to Algebra homomorphism as you already see. That is the functor we need, ${\mathrm{Top}}^{\mathrm{op}}\to \mathbb{R}-\mathrm{Algebra}$ $\mathrm{Top}^{\mathrm{op}} \to \mathrm{\mathbb R-\mathrm{Algebra}}$ .

Thus $C(a,b)$ $C{(a,b)}$ is not Noetherian too, since $(a,b)\cong \mathbb{R}\Rightarrow C(a,b)\cong C(\mathbb{R})$ $(a,b)\cong\mathbb R\Rightarrow C{(a,b)}\cong C(\mathbb R)$ .

(Why? Consider $\tan (x):(-\frac{\pi }{2},\frac{\pi }{2})\to \mathbb{R},\arctan x:\mathbb{R}\to (-\frac{\pi }{2},\frac{\pi }{2})$ $\tan(x):(-\frac{\pi}{2},\frac{\pi}{2})\to\mathbb R,\arctan x:\mathbb R\to(-\frac{\pi}{2},\frac{\pi}{2})$ , and easy to construct homeomorphism.

By the way, you have lots of choice for the target and source of the functor.

Lattice and Boolean Algebra, Algebra-Geometry duality

Consider a subset $S\subseteq \mathbb{R}$ $S\sube\mathbb R$ , $S⟼I_S:=\{f\in C(\mathbb{R})|\mathrm{\forall }x\in S,f(x)=0\}$ $S\longmapsto I_{S}:=\{f\in C(\mathbb R)|\forall x\in S,f(x)=0\}$ gives a Lattice isomorphism.

$†.S_i\subseteq S_j⟺I_{S_i}\supseteq I_{S_j}$ $\dagger .S_i\subseteq S_j\iff I_{S_i}\supseteq I_{S_j}$

For example, $\mathrm{\varnothing }⟼C(\mathbb{R})$ $\empty\longmapsto C(\mathbb R)$ , and it is a good reason for $I_{\{a\}}$ $I_{\{a\}}$ is maximal ideal.

$\begin{array}{c}†.\bigcup _{i\in I}{S}_i⟼\bigcap _{i\in I}{I}_{S_i}\end{array}$ $\dagger.\bigcup_{i\in I} S_i\longmapsto\bigcap_{i\in I} I_{S_i}\\ $

$\begin{array}{c}†.\bigcap _{i\in I}{S}_i⟼\sum _{i\in I}{I}_{S_i}\end{array}$ $\dagger.\bigcap_{i\in I} S_i\longmapsto\sum_{i\in I} I_{S_i}\\ $

And easy to see that ${\mathrm{Spec}}_{\max }(C(X))$ $\mathrm{Spec_{max}}(C(X))$ correspond to each point of $X$ $X$

Remember $C(\mathbb{R})$ $C(\R)$ is a algebra, thus those ideal is sub linear space.

If we conisder $C[a,b]$ $C{[a,b]}$ , then we can define inner product as $\begin{array}{c}⟨f,g⟩:={\int }_a^{b}f(x)g(x)dx\end{array}$ $\langle f,g\rangle:=\int_{a}^b f(x)g(x)dx\\ $

Then we have.

$†.S^c⟼I_{S^c}=(I_S{)}^{\mathrm{\perp }}$ $\dagger. S^c\longmapsto I_{S^c} =(I_S)^{\bot}$

Since $\begin{array}{c}f\in I_S,g\in I_{S^c},{\int }_a^{b}f(x)g(x)dx=0\end{array}$ $f\in I_S,g\in I_{S^c},\int_a^b f(x)g(x) dx=0\\ $

Then De Morgan Law is

$†.(A\cup B{)}^c=A^c\cap B^c⟼(I_A\cap I_B{)}^{\mathrm{\perp }}=I_A^{\mathrm{\perp }}+I_B^{\mathrm{\perp }}$ $\dagger.(A\cup B)^c=A^c\cap B^c\longmapsto (I_A\cap I_B)^{\bot}= I_A^{\bot}+I_B^{\bot}$

$†.(A\cap B{)}^c=A^c\cup B^c⟼(I_A+I_B{)}^{\mathrm{\perp }}=I_A^{\mathrm{\perp }}\cap I_B^{\mathrm{\perp }}$ $\dagger. (A\cap B)^c= A^c\cup B^c\longmapsto (I_A+I_B)^{\bot}=I_A^{\bot}\cap I_B^{\bot}$

Math Essays: Lattice and Boolean Algebra over vector space (1) (wuyulanliulongblog.blogspot.com)

You can see the general result here.

By the way, the anti isomorphism will form a Galois connection.

This idea will connect with lots of deep branch, I will write a new article about it after I learn and build enough.