# Yoneda argument – Application

A simple application of Yoneda argument is to prove that $a^1 \cong a$ in Cartesian closed category.

Lemma. In Cartesian closed category, $a^1 \cong a$ where $1$ is terminal.

Proof. $hom(b, a) \cong hom(b \times 1, a) \cong hom(b, a^1)$ for all objects $b$. Hene $a \cong a^1$.

It’s also quite easy to see that $a^{b \times c} \cong (a^b)^c$ since $hom(d, a^{b \times c}) \cong hom(d \times b \times c, a) \cong hom(d \times c, a^b) \cong hom(d, (a^b)^c)$.  Hence $a^{b \times c} \cong (a^b)^c$. It’s just like that we prvoe the equality $c \wedge b \rightarrow a = c \rightarrow (b \rightarrow a)$ in the Heyting algebra!