The Naturality of the Power Map: A Yoneda Perspective（自成天然之趣，不烦人事之工。）

When I first began my study of mathematics, I felt dissatisfied with the definition $n^0=1$$n^0 = 1$ in Tao's Analysis, which seemed like a mere convention. Why should there be such arbitrary prescriptions in mathematics?

Later, I realized that one could categorify $\mathbb{N}$$\mathbb{N}$ via the category of finite sets, where the story simply boils down to the empty set acting as the initial object. Eventually, I came to understand that many familiar operations are actually intrinsic structures naturally carried by certain algebraic categories. For instance, exponentiation arises canonically from $\mathsf{Mon}$$\mathsf{Mon}$, just as polynomial operations arise naturally from $\mathsf{Ring}$$\mathsf{Ring}$.

Let us consider the category of monoids, denoted by $\mathsf{Mon}$$\mathsf{Mon}$.

The forgetful functor is represented by $\mathbb{N}$$\mathbb{N}$.

Remark. For any Lawvere Theory , the forgetful functor is represented by $F(1)$$F(1)$, where $F$$F$ is the free functor.

The Yoneda lemma tells us that

$$\mathbb{N}\cong {\mathrm{Hom}}_{\mathsf{Mon}}(\mathbb{N},\mathbb{N})\cong \mathrm{Nat}(h_{\mathbb{N}},h_{\mathbb{N}})$$

The Yoneda lemma implies that each $n$$n$ defines a natural transformation. It is not hard to see that this natural transformation is simply the power map $(-{)}^n$$(-)^n$, and the natural transformation defined by $0$$0$ sends everything to the identity element $e$$e$.

Also, Yoneda Lemma tells us that $(x^m{)}^n=x^{mn}$$(x^{m})^n=x^{mn}$ directly.

Let us consider $m,n\in \mathbb{N}$$m,n\in\mathbb N$. Then we have $m\cdot n=mn$$m\cdot n=mn$. (For example, $6\cdot 7=42)$$6\cdot 7=42)$ Hence, after Yoneda embedding, we have

$$(-{)}^n\circ (-{)}^m=(-{)}^{mn}.$$

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Similarly, in the category of rings (specifically commutative rings), denoted by $\mathsf{Ring}$$\mathsf{Ring}$, we replace $\mathbb{N}$$\mathbb{N}$ with the polynomial ring $\mathbb{Z}[x]$$\mathbb{Z}[x]$.

Here, the forgetful functor is represented by $\mathbb{Z}[x]$$\mathbb{Z}[x]$.

By the Yoneda lemma, we have:

$$\mathbb{Z}[x]\cong {\mathrm{Hom}}_{\mathsf{Ring}}(\mathbb{Z}[x],\mathbb{Z}[x])\cong \mathrm{Nat}(h_{\mathbb{Z}[x]},h_{\mathbb{Z}[x]})$$

The natural transformation defined by a polynomial $f(x)\in \mathbb{Z}[x]$$f(x) \in \mathbb{Z}[x]$ is precisely the evaluation map. For any element $r$$r$ in a ring, the transformation is simply:

$$r⟼f(r)$$

Just as exponentiation is the canonical operation in monoids, polynomial evaluation is the canonical operation in rings.

Similarly, $0\in \mathbb{Z}[x]$$0\in\mathbb Z[x]$ gives us $r⟼0$$r\longmapsto 0$.