# Proving exponential arithmetic rules using the Yoneda lemma

Working in a Cartesian closed category we have have an exponential object $$X^Y$$ for object $$X$$ and $$Y$$. There are isomorphisms $$1^X \cong 1$$, $$X^Y \times X^Z \cong X^{Y + Z}$$, $$X^1 \cong X$$ and a few others, just like with sets. Now nlab claims they can be proven by the usual Yoneda arguments. I don't understand how to do this. See also this related question. I don't understand how this works. Let us look at that answer first.

By definition of a power object, there is a bijection between $$\operatorname{Hom}(x, a^1)$$ and $$\operatorname{Hom}(x \times a, 1)$$. Since $$\operatorname{Hom}(x \times a, 1)$$ contains just one element, there is a bijection $$\operatorname{Hom}(x \times a, 1) \cong \operatorname{Hom}(x, a)$$ as well, so $$\operatorname{Hom}(x, a^1) \cong \operatorname{Hom}(x, a)$$. I don't see how we can apply Yoneda to conclude there is an isomorphism between $$a$$ and $$a^1$$.

By Yoneda, we know that if $$x,y \in \mathsf{C}$$ are objects in some locally small category, then $$x \simeq y$$ if and only if the functors represented by these objects are naturally isomorphic (that is, either $$\mathsf{C}(x,-) \simeq \mathsf{C}(y,-)$$ or $$\mathsf{C}(-,x) \simeq \mathsf{C}(-,y)$$).

Let's apply that to prove that $$a^1 \simeq a$$. By Yoneda, it suffices to note that

$$\mathsf{C}(x,a^1) \simeq \mathsf{C}(x \times 1,a) \simeq \mathsf{C}(x,a)$$

for any $$x$$, and these isomorphisms assemble into a natural isomorphism

$$\mathsf{C}(-,a^1) \simeq \mathsf{C}(-,a)$$

which shows that $$a^1 \simeq a$$.

Likewise we have a natural isomorpshim given by the bijections

\begin{align} \mathsf{C}(c,x^y \times x^z) &\simeq \mathsf{C}(c,x^y) \times \mathsf{C}(c,x^z) \simeq \mathsf{C}(c \times y,x) \times \mathsf{C}(c \times z,x)\\ &\simeq \mathsf{C}(c \times y \sqcup c \times z,x) \simeq \mathsf{C}(c \times (y \sqcup z),x)\\ &\simeq \mathsf{C}(c,x^{y \ \sqcup \ z}). \end{align}

for each $$c \in \mathsf{C}$$, and so $$x^y \times x^z \simeq x^{y \ \sqcup \ z}$$.

You can prove the rest in a similar fashion.

• Thanks, I completely forgot about the naturality requirement of the Yoneda lemma. Oct 14, 2019 at 16:39