A very simple question on adjunctions that I simply can't look at correctly.

Question

Suppose that $\mathcal{C}$ has finite products. Why is the unit $\eta_c:c\to c\times c$ of the adjunction $\Delta\dashv \times:\mathcal{C}\times\mathcal{C}\to\mathcal{C}$ the arrow $\langle 1_c,1_c\rangle$? In other words, why is $\pi_1\circ \eta_c=1_c=\pi_2\circ\eta_c$?

Answer 1

The adjunction is saying that morphisms $$(f,g) : (c,c) \to (a,b)$$ in $\mathcal{C} \times \mathcal{C}$ correspond naturally with morphisms $$h : c \to a \times b$$ in $\mathcal{C}$. This correspondence is fairly obviously given by putting $h=\langle f,g \rangle$, where $\langle f,g \rangle$ is the unique morphism $c \to a \times b$ induced by $f : c \to a$ and $g : c \to b$ by the universal property of the product. (And vice versa.)

The $c$-component of the unit is therefore the morphism corresponding to $1_{(c,c)}$. But $1_{(c,c)} = (1_c, 1_c)$, meaning that $$\eta_c = \langle 1_c, 1_c \rangle : c \to c \times c$$ Likewise the counit is the morphism corresponding with $1_{a \times b} : a \times b \to a \times b$. But $1_{a \times b} = \langle \pi_1, \pi_2 \rangle$, and so $$\varepsilon_{(a,b)} = (\pi_1, \pi_2) : (a \times b, a \times b) \to (a,b)$$