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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 $a^1$.

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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.

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

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