The Naturalness of Addition and Multiplication: A Categorical Perspective

The aim of this blog is to explain why the $+$$+$ and $\times$$\times$ operators on $\mathbb{N}$$\mathbb{N}$ are natural from a categorical point of view.

Let us consider the category of finite sets, $\mathrm{Fset}$$\mathrm{Fset}$.

It has an initial object $0:=\mathrm{\varnothing }$$0 := \emptyset$ and a final object $1:=\{\ast \}$$1 := \{\ast\}$, finite products $X\times Y$$X \times Y$, and coproducts $X+Y$$X + Y$, exponential object $Y^X$$Y^X$. Here $Y^X:={\mathrm{Hom}}_{\mathrm{Fset}}(X,Y)$$Y^X := \mathrm{Hom}_{\mathrm{Fset}}(X, Y)$.

It is not hard to see the correspondence between $\mathrm{Fset}$$\mathrm{Fset}$ and $\mathbb{N}$$\mathbb{N}$. I would like to say that $+$$+$ and $\times$$\times$ are the most natural two binary operations you can define. Why? If you want to define a binary operator on a set $S$$S$ or something similar, you have to pass through $\mathrm{\Delta }(S)=S\times S$$\Delta(S) = S \times S$. Since a binary operator is a function from $S\times S$$S \times S$ to $S$$S$.

The next proposition will claim that $+$$+$ and $\times$$\times$ actually come from the adjoint of $\mathrm{\Delta }$$\Delta$.

Proposition. $+⊣\mathrm{\Delta }⊣\times$$+ \dashv \Delta \dashv \times$. Here, $\mathrm{\Delta }:\mathcal{C}\to \mathcal{C}\times \mathcal{C},X⟼X\times X$$\Delta: \mathcal{C} \to \mathcal{C} \times \mathcal{C}, X \longmapsto X \times X$ is the diagonal functor. Coproduct and product are binary functors from $\mathcal{C}\times \mathcal{C}$$\mathcal{C} \times \mathcal{C}$ to $\mathcal{C}$$\mathcal{C}$, if the coproduct and product exist in $\mathcal{C}$$\mathcal{C}$.

Proof.

We need to prove that:

$$\begin{array}{}\text{(1)}& {\mathrm{Hom}}_{\mathcal{C}}(X+Y,Z)\cong {\mathrm{Hom}}_{\mathcal{C}\times \mathcal{C}}((X,Y),\mathrm{\Delta }(Z))\end{array}$$

and

$$\begin{array}{}\text{(2)}& {\mathrm{Hom}}_{\mathcal{C}\times \mathcal{C}}(\mathrm{\Delta }(X),(Y,Z))\cong {\mathrm{Hom}}_{\mathcal{C}}(X,Y\times Z)\end{array}$$

Both follow directly from the universal property (draw the diagram then see). $◻$$\square$

This adjunction tells us that in the category of finite sets,

$$\begin{array}{}\text{(3)}& Z^{X+Y}\cong Z^X\times Z^Y,Y^X\times Z^X\cong (Y\times Z{)}^X\end{array}$$

Since if $\mathcal{C}$$\mathcal{C}$ is a locally small category, then the target of ${\mathrm{Hom}}_{\mathcal{C}}(-,-)$$\mathrm{Hom}_{\mathcal{C}}(-, -)$, ${\mathrm{Hom}}_{\mathcal{C}\times \mathcal{C}}(-,-)$$\mathrm{Hom}_{\mathcal{C} \times \mathcal{C}}(-, -)$ are $\mathrm{Set}$$\mathrm{Set}$!

We also have the product-exponential adjunction (or tensor-hom adjunction if you view $\mathrm{Fset}$$\mathrm{Fset}$ as a tensor category):

$$\begin{array}{}\text{(4)}& {\mathrm{Hom}}_{\mathrm{Fset}}(X\times Y,Z)\cong {\mathrm{Hom}}_{\mathrm{Fset}}(X,Z^Y)\end{array}$$

Hence we have $Z^{X\times Y}\cong (Z^Y{)}^X$$Z^{X \times Y} \cong (Z^Y)^X$.

Obviously, $\times$$\times$ and $+$$+$ are associative and commutative by the property of product and coproduct.

The distributive law:

$$\begin{array}{}\text{(5)}& (A+B)\times Y\cong A\times Y+B\times Y\end{array}$$

holds since $-\times Y$$- \times Y$ is a left adjoint of ${\mathrm{Hom}}_{\mathrm{Fset}}(Y,-)$$\mathrm{Hom}_{\mathrm{Fset}}(Y, -)$ and hence preserves colimits.

The natural isomorphism ${\mathrm{Id}}_{\mathrm{Fset}}\cong {\mathrm{Hom}}_{\mathrm{Fset}}(\{\ast \},-)$$\mathrm{Id}_{\mathrm{Fset}} \cong \mathrm{Hom}_{\mathrm{Fset}}(\{\ast\}, -)$ tells us that $Y^1\cong Y$$Y^1 \cong Y$. $\{\ast \}$$\{\ast\}$ being the final object tells us that $1^X\cong 1$$1^X \cong 1$.

The empty set $0$$0$ is the initial object, hence $Y^0\cong 1$$Y^0 \cong 1$. There is no function $f:X\to \mathrm{\varnothing }$$f: X \to \emptyset$. Hence $0^X\cong 0$$0^X \cong 0$.

Obviously, $\times$$\times$ and $+$$+$ are associative and commutative by the property of products and coproducts.

The distributive law:

$$\begin{array}{}\text{(6)}& (A+B)\times Y\cong A\times Y+B\times Y\end{array}$$

holds since $-\times Y$$-\times Y$ is a left adjoint of ${\mathrm{Hom}}_{\mathrm{Fset}}(Y,-)$$\mathrm{Hom}_{\mathrm{Fset}}(Y, -)$ and hence preserves colimits.

The natural isomorphism ${\mathrm{Id}}_{\mathrm{Fset}}\cong {\mathrm{Hom}}_{\mathrm{Fset}}(\{\ast \},-)$$\mathrm{Id}_{\mathrm{Fset}} \cong \mathrm{Hom}_{\mathrm{Fset}}(\{\ast\}, -)$ tells us that $Y^1\cong Y$$Y^1 \cong Y$. $\{\ast \}$$\{\ast\}$ being the final object tells us that $1^X\cong 1$$1^X \cong 1$.

The empty set $0$$0$ is the initial object, hence $Y^0\cong 1$$Y^0 \cong 1$. There is no function $f:X\to \mathrm{\varnothing }$$f: X \to \emptyset$. Hence $0^X\cong 0$$0^X \cong 0$​.

$0\times X\cong 0$$0\times X\cong 0$ since ${\mathrm{Hom}}_{\mathrm{Fset}}(0\times X,Y)\cong {\mathrm{Hom}}_{\mathrm{Fset}}(0,Y^X)$$\mathrm{Hom}_{\mathrm{Fset}}(0\times X,Y)\cong\mathrm{Hom}_{\mathrm{Fset}}(0,Y^X)$, and there exists a unique homomorphism from $0$$0$ to $Y^X$$Y^X$ since $0$$0$ is the initial object. Hence $\mathrm{\forall }Y,\mathrm{\exists }!f:0\times X\to Y$$\forall Y,\exists !f:0\times X\to Y$. Hence $0\times X$$0\times X$ is initial object as well. Hence $0\times X\cong 0$$0\times X\cong 0$​.

Similarly, $X\times 1\cong X$$X\times 1\cong X$ since ${\mathrm{Hom}}_{\mathrm{Fset}}(X\times 1,Y)\cong {\mathrm{Hom}}_{\mathrm{Fset}}(X,Y^1)\cong {\mathrm{Hom}}_{\mathrm{Fset}}(X,Y)$$\mathrm{Hom}_{\mathrm{Fset}}(X\times 1,Y)\cong \mathrm{Hom}_{\mathrm{Fset}}(X,Y^1)\cong \mathrm{Hom}_{\mathrm{Fset}}(X,Y)$. By Yoneda lemma, $X\times 1\cong X$$X\times 1\cong X$.

As you can see, the adherence of these arithmetic laws is not due to magic or coincidence. The operations $+$$+$, $\times$$\times$, and even the exponential are defined by the adjoint functor. These arithmetic laws either stem from the isomorphism of two $\mathrm{Hom}$$\mathrm{Hom}$ functors or from the properties of the adjoint functor, i.e., their ability to preserve colimits. The specific values of $Y^X$$Y^X$ are determined by the universal properties of initial and final objects. The cardinality gives us the natural number semantic meaning. But, as you can see, lots of the argument do not depend on the properties of $\mathrm{F}\mathrm{s}\mathrm{e}\mathrm{t}$$\rm Fset$. We will generalize it to cartesian closed category dive in to topos theory.