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!

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