Counting with Natural Numbers: A Categorical Perspective

 

The idea is to use category theory to define $+,\times ,Y^X,2^X$$+,\times,Y^X,2^X$ and prime number in $\mathbb{N}.$$\mathbb N.$

Just for fun. Do not use category theory to teach six-year-old kids to count.

Consider $\mathrm{FinSet}$$\mathrm{FinSet}$ , then

$†.\mathrm{\varnothing }$$\dagger.\empty$is set.

$†.$$\dagger.$ if $S$$S$ is a Finite set, then $S\coprod \{\ast \}$$S\coprod\{*\}$ is a finite set.

$†.$$\dagger.$ for any finite set $S$$S$, $S\coprod \{\ast \}\ncong \mathrm{\varnothing }$$S\coprod\{*\}\ncong\empty$.

$†.ForN,M\in \mathrm{FinSet}$$\dagger. For N,M\in \mathrm{FinSet}$, if $N\coprod \{\ast \}\cong M\coprod \{\ast \}$$N\coprod\{*\}\cong M\coprod\{*\}$, then $N\cong M$$N\cong M$

$†.$$\dagger.$Let $P(N)$$P(N)$ represent a proposition of $N$$N$ ($N$$N$ up to isomorphism),

if $P(\mathrm{\varnothing })$$P(\empty)$ is true, $P(N)$$P(N)$ is true implies $P(N\coprod \{\ast \})$$P(N\coprod\{*\})$ is true, then for all finite set $M$$M$, $P(M)$$P(M)$ Is true.

+

Traditionally, we need to define

$0+n=n.$$0+n=n.$

$(n++)+m=(n+m)++$$(n++)+m=(n+m)++$

But now, $n+m$$n+m$ could be defined as $N\coprod M$$N\coprod M$.

$N=\{1,2,3,...,n\},M=\{1,2,3,...,m\},N\coprod M\cong \{1,2,3,...,n+m\}$$N=\{1,2,3,...,n\},M=\{1,2,3,...,m\},N\coprod M\cong\{1,2,3,...,n+m\}$.

It is associative, commutative, and has cancel law. $(A\coprod B\cong A\coprod C\Rightarrow B\cong C)$$(A\coprod B\cong A\coprod C\Rightarrow B\cong C)$

We will use the notation $A+B$$A+B$ for the coproduct.

$\times$$\times$

Traditionally, we need to define

$0\times m=0,(n++)\times m=n\times m+m$$0\times m=0,(n++)\times m=n\times m+m$

But now, we need to define $N\times M$$N\times M$. Again, it is associative, commutative, and has the cancel law.

Easy to see $N\times M\cong \{1,2,3,...,nm\}$$N\times M\cong\{1,2,3,...,nm\}$.

$Y^X$$Y^X$

Traditionally...

But now, we only need to define $Y^X:={\mathrm{Hom}}_{\mathrm{FinSet}}(X,Y).$$Y^X:=\mathrm{Hom}_{\mathrm{FinSet}}(X,Y).$

Then you will see that $0^0=1,Y^0=1\in \mathbb{N}$$0^{0}=1,Y^0=1 \in \mathbb N$. Since $\mathrm{\varnothing }$$\emptyset$ is the initial object, for any finite set, there exists a unique arrow.

Easy to see $Y^X\cong \{1,2,3,...,Y^X\}$$Y^X\cong\{1,2,3,...,Y^X\}$.

Remark

In fact, what we do here is just consider a functor, map the gropoid to one point.

That is, $F(\mathrm{\varnothing })=0,F(\{\ast \})=1)...$$F(\empty)=0,F(\{*\})=1)...$

$2^X$$2^X$

Now you can define $2^X:={\mathrm{Hom}}_{\mathrm{FinSet}}(X,2)$$2^X:=\mathrm{Hom}_{\mathrm{FinSet}}(X,2)$. In this case, $2$$2$ refers to ${\mathbb{F}}_2$$\mathbb F_2$.

You may have seen something interesting.

Like in $\mathrm{FinSet}$$\mathrm{FinSet}$, $\mathrm{\varnothing }$$\empty$ is the initial object and $\{\ast \}$$\{*\}$ is the final object and the duality between Product and Coproduct.

Then, we can consider the ${\mathrm{FinSet}}^{\mathrm{o}\mathrm{p}}$$\mathrm{FinSet}^{\rm op}$. But an exciting result we know is ${\mathrm{Set}}^{\mathrm{o}\mathrm{p}}\cong \mathrm{Bool}$$\mathrm{Set}^{\rm op}\cong\mathrm{Bool}$.

$\mathrm{Bool}$$\mathrm{Bool}$ is the category of complete and atomic Boolean Algebra.

The initial idea is $f^{-1}$$f^{-1}$ preserve $\cup ,\cap {,}^c$$\cup,\cap,^c$, thus it prserve symmetry difference. Therefore, it is a Boolean Ring homomorphism.

But now, you can consider this functor ${\mathrm{Hom}}_{\mathrm{FinSet}}(-,2)$$\mathrm{Hom}_{\mathrm{FinSet}}(-,2)$.

$(f:X\to Y)⟼(f^{\ast }:{\mathrm{Hom}}_{\mathrm{FinSet}}(Y,2)⟶{\mathrm{Hom}}_{\mathrm{FinSet}}(X,2))$$(f:X\to Y)\longmapsto(f^*:\mathrm{Hom}_{\mathrm{FinSet}}(Y,2)\longrightarrow\mathrm{Hom}_{\mathrm{FinSet}}(X,2))$, then consider the functor ${\mathrm{Hom}}_{\mathrm{FinSet}}(X,2)\to P(X),f^{\ast }\to f^{-1}$$\mathrm{Hom}_{\mathrm{FinSet}}(X,2)\to P(X),f^*\to f^{-1}$.

It is equivalent, $Y_i\subseteq Y$$Y_i\subseteq Y$ is just a function about each $y_i$$y_i$ exists or not.

Then $f^{\ast }(Y_i)(x)=(Y_i\circ f)(x)=1⟺x\in f^{-1}(Y_i)$$f^*(Y_i)(x)=(Y_i\circ f)(x)=1\iff x\in f^{-1}(Y_i)$.

For every finite complete atomic Boolean Algebra $\mathcal{B}$$\mathcal B$ which atom set is $\mathcal{A}$$\mathcal A$, you can always consider $b⟼\{a\in \mathcal{A}|a\le b\}$$b\longmapsto\{a\in\mathcal A|a\le b\}$.

This is a Boolean isomorphism $\mathcal{B}\cong P(\mathcal{A})$$\mathcal B\cong P(\mathcal A)$. This result is true in $\mathrm{S}\mathrm{e}\mathrm{t}$$\rm Set$ as well.

That is the functor from $\mathrm{B}\mathrm{o}\mathrm{o}\mathrm{l}\to \mathrm{S}\mathrm{e}\mathrm{t}$$\rm Bool\to Set$. ${\log }_2(\mathcal{B})=\mathcal{A}$$\log_2(\mathcal B)=\mathcal A$ .

How does this anti-isomorphism work for the duality of Product and Coproduct?

It is pretty easy! Consider the coproduct; it just $2^{\mathrm{X}+\mathrm{Y}}=2^X\times 2^Y$$2^{\mathrm{X+Y}}=2^X\times 2^{Y}$!

Or ${\log }_2({\mathcal{B}}_n\times {\mathcal{B}}_m)={\log }_2({\mathcal{B}}_n)+{\log }_2({\mathcal{B}}_m)$$\log_2(\mathcal B_n\times \mathcal B_m)=\log_2(\mathcal B_n)+\log_2(\mathcal B_m)$ .

By the way, $f^{\ast }:2^Y\to 2^X,{\log }_2(f^{\ast })=f$$f^*:2^Y\to2^X,\log_2(f^*)=f$, since you can view $f^{\ast }\subseteq 2^Y\times 2^X$$f^*\sube 2^Y\times 2^X$ .

Then for $f^{\ast }(\{y\})=(\{y\},\{x_1,x_2,...,x_n\}),{\log }_2(f^{\ast }(\{y\}))=\{y,x_1,x_2,x_3,...,x_n\}$$f^*(\{y\})=(\{y\},\{x_1,x_2,...,x_n\}),\log_2(f^*(\{y\}))=\{y,x_1,x_2,x_3,...,x_n\}$.

If you consider the product, $2^{X\times Y}=(2^X{)}^Y$$2^{X\times Y}=(2^{X})^Y$. That is, ${\mathrm{Hom}}_{\mathrm{FinSet}}(X\times Y,2)\cong {\mathrm{Hom}}_{\mathrm{FinSet}}(Y,2^X)\cong {\mathrm{Hom}}_{\mathrm{FinSet}}(X,2^Y)$$\mathrm{Hom}_{\mathrm{FinSet}}(X\times Y,2)\cong\mathrm{Hom}_{\mathrm{FinSet}}(Y,2^X)\cong\mathrm{Hom}_{\mathrm{FinSet}}(X,2^Y)$.

This does form a Boolean Algebra.

Since Boolean Algebra is equivalence to Boolean Ring and $2^X$$2^X$ is isomorphic to the product of ${\mathbb{F}}_2$$\mathbb F_2$.

You can check the universal property of the coproduct for $2^{X\times Y}$$2^{X\times Y}$.

$i_X:=2^X\to 2^{X\times 0_Y},i_Y:=2^Y\to 2^{0_X\times Y}$$i_X:=2^X\to 2^{X\times{0_Y}},i_Y:=2^Y\to 2^{0_X\times Y}$.

Then, for any Boolean ring $Z$$Z$,

the unique homomorphism from $f:2^{X\times Y}\to Z$$f:2^{X\times Y}\to Z$ makes the diagram commute is $f(x,y)=f_X(x,0)+f_Y(0,y)$$f(x,y)=f_X(x,0)+f_Y(0,y)$.

An interesting idea

In its contemporary form, a “geometric object” is usually defined as an “object” that “locally” “looks like” a “standard geometric object”. Depending on the geometry that one is interested in, there will be very different “standard geometric objects” as the basic building blocks. For the theory of (finite-dimensional) manifolds one chooses open subsets of finite-dimensional R- or C-vector spaces together with their “differentiable structure”. To make the notion of a geometric object precise, one proceeds in general as follows. The first one introduces the language of categories yielding the notions of objects and the precise meaning of “looks like” as being isomorphic in that category. Next one has to find a (maybe very large) category C that contains the “standard geometric objects” as a subcategory and in which it makes sense to use the word “locally”. Then, finally one can give the precise definition of a geometric object as an object of C that is locally isomorphic to an object in the subcategory of standard geometric objects.-------Manifolds, Sheaves, and Cohomology

If you let $N=\{1,2,3,,...,n\}...$$N=\{1,2,3,,...,n\}...$ be the standard object, and let us consider $\mathrm{Set}$$\mathrm{Set}$.

Every set locally looks like a finite set $\{1,2,3...,n\}$$\{1,2,3...,n\}$, that is, for every set, there exists a subset isomorphic to $N$$N$.

We may need AC here.

You might ask how to define the usual order of $\mathbb{N}$$\mathbb N$ here?

The functor $F$$F$ will map inclusion to $\le$$\le$ !

Now, if you only consider the inclusion map to be the morphism, then

Presheaf

Consider ${\mathrm{Hom}}_{\mathrm{FinSet}}(-,2)$$\mathrm{Hom}_{\mathrm{FinSet}}(-,2)$, it maps each finite set to the Boolean Ring $(P(S),\mathrm{\Delta },\cap )$$(P(S),\Delta,\cap)$.

$(i:X\to Y)⟶(i^{\ast }:(P(Y),\mathrm{\Delta },\cap )\to (P(X),\mathrm{\Delta },\cap ))$$(i:X\to Y)\longrightarrow(i^*:(P(Y),\Delta,\cap)\to(P(X),\Delta,\cap))$.

More naturally, consider the image of the functor $F$$F$, that is $(\mathbb{N},i)$$(\mathbb N,i)$.

$N\in {\mathrm{Ob}}_{\mathbb{N}}$$N\in\mathrm{Ob}_{\mathbb N}$ is $N=\{1,2,3,...,n\}$$N=\{1,2,3,...,n\}$(maybe you can consider $\mathbb{Z}/m\mathbb{Z}$$\mathbb Z/m\mathbb Z$).

It is just the Boolean Ring Presheaf over $\mathrm{FinSet}$$\mathrm{FinSet}$, or more generally, $\mathrm{Set}$$\mathrm{Set}$.

We could have a geometry view of Set theory.

Like, what is the kernel of $i^{\ast }:(P(M),\mathrm{\Delta },\cap )\to (P(N),\mathrm{\Delta },\cap )?$$i^{*}:(P(M),\Delta,\cap)\to(P(N),\Delta,\cap)?$

The kernel is $2^{M-N}=2^{\{n+1,...,m\}}\cong 2^{\{1,2,3,...,m-n\}}$$2^{M-N}=2^{\{n+1,...,m\}}\cong2^{\{1,2,3,...,m-n\}}$ .

In general, every stone space has a Boolean Ring presheaf.

You can consider the open set category and give each open set subspace topology.

Then each open set is stone space as well.

We will define $M-N=\{1,2,3,...,m-n\}$$M-N=\{1,2,3,...,m-n\}$ if $\mathrm{\exists }i:N\to M$$\exists i:N\to M$.

Or, we can write it as $2^{M-N}=\frac{2^M}{2^N}$$2^{M-N}=\frac{2^{M}}{2^N}$, and it is just the cokernel of $i:(P(N),\mathrm{\Delta },\cap )\to (P(M),\mathrm{\Delta },\cap )$$i:(P(N),\Delta,\cap)\to (P(M),\Delta,\cap)$.

Since we can consider the isomorphism $P(K)\cong {\mathbb{F}}_2^K$$P(K)\cong\mathbb F_2^K$, thus you can view it as a ${\mathbb{F}}_2-\text{Vector Space}$$\mathbb F_2-\text{Vector Space}$.

Homology?

In fact, ${\mathbb{F}}_2-\text{Algebra}$$\mathbb F_2-\text{Algebra}$. Then $\begin{array}{c}\frac{2^M}{2^N}\end{array}$$\frac{2^M}{2^N}\\$ is just the $\begin{array}{c}\frac{\mathrm{cod}i}{\mathrm{im}i}\end{array}$$\frac{\mathrm{cod}i}{\mathrm{im}i}\\$.

If we consider

$0\stackrel{i_0}{\to }N\stackrel{i_N}{\to }M\stackrel{i_M}{\to }K$$0\xrightarrow{i_0}N\xrightarrow{{i_N}} M\xrightarrow{i_{M}} K$

Then

$2^K\stackrel{i_M^{\ast }}{\to }{2}^M\stackrel{i_N^{\ast }}{\to }{2}^N\stackrel{i_0^{\ast }}{\to }0$$2^K\xrightarrow{i_M^*}2^M\xrightarrow{i_N^*}2^N\xrightarrow{i_0^*}0$.

Maybe we could have $0\stackrel{i_N}{\to }N\stackrel{i_{N\times M}}{\to }N\times M\stackrel{i_M}{\to }M\stackrel{i_0}{\to }0$$0\xrightarrow{i_N} N\xrightarrow{i_{N\times M}}N\times M\xrightarrow{i_M}M\xrightarrow{i_0}0$ ? in the sense of vector space.

$0⟶2^K\stackrel{i_K}{\to }{2}^M\stackrel{i_N^{\ast }}{\to }{2}^N⟶0$$0\longrightarrow2^K\xrightarrow{i_K}2^M\xrightarrow{i_N^*}2^N\longrightarrow 0$, then $\mathrm{Ker}(i_N^{\ast })=2^{M-N}=2^K$$\mathrm{Ker}(i_N^*)=2^{M-N}=2^K$,

If $M-N\ge K$$M-N\ge K$, then $i_N^{\ast }\circ i_K=0$$i^*_N\circ i_K=0$ , it is closed, since $\mathrm{Ker}(i_N^{\ast })\supseteq \mathrm{Im}(i_K)$$\mathrm{Ker}(i^*_N)\supseteq\mathrm{Im}(i_K)$

If $M-N=K$$M-N=K$, then $i_N^{\ast }\circ i_K=0$$i_N^*\circ i_K=0$ , it is exact, since $\mathrm{Ker}(i_N^{\ast })=\mathrm{Im}(i_K)$$\mathrm{Ker}(i^*_N)=\mathrm{Im}(i_K)$.

However, I have not learned homology and cohomology. I will back to it after I learn it.

Prime Number

we will see a prime number if $P\cong A\times B$$P\cong A\times B$, then $A\cong \{\ast \}$$A\cong\{*\}$, $B\cong P$$B\cong P$, or $A\cong P,B\cong \{\ast \}$$A\cong P,B\cong\{*\}$.

Fundamental Theorem of Arithmetic

Every Finite set can be written as $M\cong P_{i_1}^{m_1}\times P_{i_2}^{m_2}\times ...\times P_{i_j}^{m_j}$$M\cong P_{i_1}^{m_1}\times P_{i_2}^{m_2}\times...\times P_{i_j}^{m_j}$

You can define Euclidean division, $A\equiv B\mathrm{mod}M$$A\equiv B\mod M$ as well.

For example, $\{a,b,c\}\equiv \{\mathbb{N},\mathbb{Z},\mathbb{Q},\mathbb{R},\mathbb{C}\}\mathrm{mod}\{1,2\}$$\{a,b,c\}\equiv\{\mathbb N,\mathbb Z,\mathbb Q,\mathbb R,\mathbb C\}\mod \{1,2\}$.

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