Confusion about (part of) the definition of unit and counit of adjunction

Question

Following p.51 from here:

The set-up is two functors $F:\mathscr A\to\mathscr B,G:\mathscr B\to \mathscr A$ and $F\dashv G$. So there is a bijection $$\mathscr B(F(A),B)\cong\mathscr A(A,G(B))\\g\mapsto\overline g$$ and the inverse is written as $f\mapsto \overline f$.

For each $A\in\mathscr A$, we have a map $$(\eta_A:A\to GF(A))=(\overline{1:F(A)\to F(A)}).$$ Dually, for each $B\in\mathscr B$, we have a map $$(\epsilon_B: FG(B)\to B)=(\overline{1:G(B)\to G(B)}).$$

1) Where do these equalities come from?

2) Does the term "dually" have a precise meaning? Informally, dualizing reverses all arrows. But in this case it's not reversing the arrows ($\epsilon_B$ is not $\eta_A$ with arrow reversed (which would be $GF(A)\to A)$, it's something different).

Answer 1

Here's how I understand the situation.

1) Set $B=F(A)$ (in the notation of the set-up). Then there is a bijection $$\mathscr B(F(A),F(A))\cong \mathscr A(A,GF(A))$$ which sends, in particular, $1_{F(A)}:F(A)\to F(A)$ to $\overline {1_{F(A)}}:A\to GF(A)$. Define $\eta_A$ to be that $\overline{1_{F(A)}}$.

2) The functors $F$ and $G$ induce the dual functors $F^{op}:\mathscr A^{op}\to\mathscr B^{op}$ and $G^{op}:\mathscr B^{op}\to\mathscr A^{op}$. The functor $F^{op}$ is defined by $F^{op}(A)=F(A)$ for an object $A$ and by $F^{op}(f^{op})=F(f)^{op}$ for an arrow $f^{op}$. Since $F\dashv G$, it follows from this answer that $G^{op}\dashv F^{op}$. So there is a bijection $$\mathscr A^{op}(G(B),A)\cong \mathscr B^{op}(B,F(A))$$ for all objects $A\in\mathscr A,B\in\mathscr B$. Taking $A$ to be $G(B)$, we have a bijection $$\mathscr A^{op}(G(B),G(B))\cong \mathscr B^{op}(B,FG(B)).$$ It sends, in particular, $1_{G(B)}:G(B)\to G(B)$ to $\overline{1_{G(B)}}:B\to FG(B)$. Thus $\overline{1_{G(B)}}^{op}$ is an arrow $FG(B)\to B$. Define $\epsilon_B$ to be that $\overline{1_{G(B)}}^{op}$.

Answer 2

1)

They come from the identifications of the adjunction. Let $F \colon \mathcal{C} \rightarrow \mathcal{D}$ be a left adjoint to $G \colon \mathcal{D} \rightarrow \mathcal{C}$. As you statet that means we have a bijection $\text{Hom}_{\mathcal{D}}(F(X),Y) = \text{Hom}_{\mathcal{C}}(X,G(Y))$ which is functorial in $X \in \mathcal{C}$ and $Y \in \mathcal{D}$. In other words we have a natural isomorphism between $\text{Hom}_{\mathcal{D}}(F(-),-)$ and $\text{Hom}_{\mathcal{C}}(-,G(-))$. Applying $\text{Hom}_{\mathcal{D}}(F(-),-)$ to $(X,F(X))$ then gives a bijection $\text{Hom}_{\mathcal{D}}(F(X),F(X)) = \text{Hom}_{\mathcal{C}}(X,G(F(X)))$ and thus we get a morphism $X \rightarrow G(F(X))$ that corresponds to $\text{id}_{F(X)}$. That is exactly the first equality that you stated. Analogously, we get the second one.

2)

I guess dually here is rather meant as analogously to be honest. But one can make it a bit more precise. As I explained in 1) we have a morphism $X \rightarrow G(F(X))$ that corresponds to $\text{id}_{F(X)}$ for every $X \in \mathcal{C}$. One can now check that this is functorial which means we have a natural transformation $\text{id}_{\mathcal{C}} \rightarrow G \circ F$ which is called a unit of the adjunction. The counit is the natural transformation we get when we do all the steps dually/analogously.