Show $\operatorname{Hom}(1, [\cdot \times B \to C]) \cong \operatorname{Hom}(\cdot \times B, C)$ in a cartesian closed category?

Question

Consider a cartesian closed category where the exponential object is $[A \to B] = B^A$. The following kind of isomorphism is stated in my book as evident:

$\operatorname{Hom}(1, [\cdot \times B \to C]) \cong \operatorname{Hom}(\cdot \times B, C)$

Here $1$ should be the terminal object of the category. Why is this isomorphism evident?

related https://math.stackexchange.com/questions/148864/exponential-objects-in-a-cartesian-closed-category-a1-cong-a — user1868607, May 07 '20 at 21:39

score 3 · Accepted Answer · answered May 07 '20 at 21:48

$\mathrm{Hom}(1, A \times B \Rightarrow C) \simeq \mathrm{Hom}(1 \times (A \times B), C) \simeq \mathrm{Hom}(A \times B, C)$. The first isomorphism follows from uncurrying, following exponentiation being right-adjoint to the cartesian product. The second isomorphism follows from the terminal object being the unit (up to isomorphism) for the cartesian product.

Show $\operatorname{Hom}(1, [\cdot \times B \to C]) \cong \operatorname{Hom}(\cdot \times B, C)$ in a cartesian closed category?

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