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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).

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    $\begingroup$ You should spend more time working things out for yourself. You've asked several questions today and yesterday within the space of a few hours. Quoting Barr and Wells in "Toposes, Triples and Theories", "This means that if you want to gain a thorough understanding of the material, you should be prepared to stop every few sentences (or even every sentence) and verify the claims made there in detail. You should be warned that a statement such as, 'It is easy to see...' does not mean it is necessarily easy to see without pencil and paper!" Leinster is not as brutal as TTT, but the advice applies. $\endgroup$ – Derek Elkins left SE Jun 25 '19 at 22:48
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    $\begingroup$ The equalities in (1) are the definitions of $\eta$ and $\epsilon$. "Dually" means, as you thought, reversing all arrows, but that would make $G$ the left-adjoint rather than the right-adjoint of $F$. So to keep the roles of $F$ and $G$ as they were, the author has also interchanged these. The result is that the formula defining $\eta$ turns into the formula defining $\epsilon$. $\endgroup$ – Andreas Blass Jun 25 '19 at 23:56
  • $\begingroup$ @AndreasBlass I posted an answer that includes how I now understand "dually", but I wouldn't say that $\eta$ turns into $\epsilon$ "for free" according to what I wrote in the answer, but I may be overcomplicating things. Is my approach what you meant in your comment? $\endgroup$ – user634426 Jun 25 '19 at 23:59
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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.

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  • $\begingroup$ Hm, there is no a single word about the functor you mentioned in the text (at least up to the point from where I've taken the quote), I wonder how the reader is supposed to understand where those equalities come from? I must be missing something... $\endgroup$ – user634426 Jun 25 '19 at 23:17
  • $\begingroup$ Well... this is how I explain units and counits and other people (including authors) might just be a bit quicker defining them or present it differently. The author of your book/paper/whatever just directly made use of the adjunction to get the morphism corresponding to the identity on $F(X)$. $\endgroup$ – Con Jun 25 '19 at 23:21
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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 $\overline{1_{G(B)}}^{op}$.

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  • $\begingroup$ 1) There are some typos/formal mistakes, as you will get $G(F(A))$ and not of $B$. But otherwise I agree. $\endgroup$ – Con Jun 25 '19 at 23:51
  • $\begingroup$ 2) Yes that also works. I did not think of that in that short amount of time, but yes, thats how it really works dually. $\endgroup$ – Con Jun 25 '19 at 23:52

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