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

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 $\pi_1\circ \eta_c=1_c=\pi_2\circ\eta_c$?

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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 $$\varepsilon_{(a,b)} = (\pi_1, \pi_2) : (a \times b, a \times b) \to (a,b)$$
Hmm, thanks! I actually tried to find the explicit form of the adjunct, and, if $\phi$ is the bijection of the adjunction, and $f:c\to a$, $g:c\to b$, I found via naturality, that $$\phi(f,g)=(f\times g)\circ \eta_c$$ I struggle a bit to see why this is precisely $\langle f,g\rangle$. Am I doing something wrong? – Niels.Remb05 Apr 25 '14 at 5:38
@Niels.Remb05: Right, and now $$(f \times g) \circ \eta_c = (f \times g) \circ \langle 1_c, 1_c \rangle = \langle f \circ 1_c, g \circ 1_c \rangle = \langle f,g \rangle$$ Remember $f \times g = \langle f \circ \pi_1, g \circ \pi_2 \rangle$. – Clive Newstead Apr 25 '14 at 19:46
... but this uses the fact that $\eta_c=\langle 1_c,1_c\rangle$, which is what I want to prove, doesn't it?...I was trying to prove that form of the adjunct, in order to prove that $\eta_c=\langle 1_c,1_c\rangle$...Isn't this your argument? – Niels.Remb05 Apr 25 '14 at 19:52
So either (a) check that the isomorphism $(\mathcal{C} \times \mathcal{C})((c,c),(a,b)) \cong \mathcal{C}(c,a \times b)$ given by $(f,g) \mapsto \langle f,g \rangle$ is natural in $c$ and in $(a,b)$ and find what $1_{(c,c)}$ maps to to get your unit $\eta_c$; or (b) define $\eta_c = \langle 1_c, 1_c \rangle$ and verify that the map $(f,g) \mapsto (f \times g) \circ \eta_c$ induces a natural isomorphism on hom sets. (Likewise for the counit.) – Clive Newstead Apr 25 '14 at 20:38