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I am reading chapter 2 of Tom Leinster's Basic Category Theory. In section 2.1 he defines the naturality condition with two equations. For categories $\mathscr{A}$ and $\mathscr{B}$ and morphisms $p: A' \rightarrow A$, $f: A \rightarrow G(B)$ in $\mathscr{A}$ and $q: B \rightarrow B'$, $g: F(A) \rightarrow B$ in $\mathscr{B}$ if $F$ and $G$ are adjoint functors we have $\overline{q \circ g} = G(q) \circ \overline{g}$ and $\overline{f \circ p} = \overline{f} \circ F(p)$.

In section 2.2 he claims it follows from the naturality condition that for each $A \in \mathscr{A}$ we have a map $\eta_A : A \rightarrow GF(A)$ such that $\eta_A = \overline{1_{F(A)}}$. Dually, for each $B \in \mathscr{B}$ we have a map $\epsilon_B : FG(B) \rightarrow B$ such that $\epsilon_B = \overline{1_{G(B)}}$. Moreover, these define natural transformations.

I am struggling to see how this follows from the naturality conditions. To construct a map $A \rightarrow GF(A)$ from the given conditions it seems like I would need to set $B = F(A)$, $A = A'$ and $p = 1_A$. This approach leads nowhere as all I can get is $f = \overline{\overline{f}}$.

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  • $\begingroup$ Constructing $\eta_A$ has nothing to do with setting $p$ to identity. It just says that $\eta_A$ is defined to be image of identity morphism of $F(A)$, which in $\text{Hom}(A,GF(A))$ by adjoint condition. $\endgroup$ – user160738 Jun 18 '17 at 20:35
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The naturality of Hom bijection can be expressed via following commutative diagram:

$\require{AMScd}$ $$\begin{CD} id_{F(A)}\in Hom(F(A),F(A)) @>\text{Adjoint bijection}>> Hom(A,GF(A))\\ @V \text{postcompose}\;\; F(A\to A') VV @V \text{postcompose}\;\; GF(A\to A')VV\\ Hom(F(A),F(A')) @>\text{Adjoint bijection}>> Hom(A,GF(A'))\\ @A{\text{precompose} \;\; F(A\to A')}AA @A {\text{precompose} \;\; A\to A'}AA\\ id_{F(A')}\in Hom(F(A'),F(A')) @>\text{Adjoint bijection}>> Hom(A',GF(A')) \end{CD}$$

Upward and downward directions each represent one part of axiom.

The thing here to notice is that under vertical maps, both $id_{F(A)}$ and $id_{F(A')}$ map to same morphism in $Hom(F(A),F(A'))$ (this is essentially by axiom on identity morphisms).

So their end-image in $Hom(A,GF(A'))$ must conincide when following arrows vertically then horizontally.

But by commutativity of diagram, then by definition of $\eta$ one must have

$$ GF(A\to A')\circ\eta_A=\eta_{A'} \circ (A\to A') $$

which is exactly the naturality of $\eta$

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  • $\begingroup$ Thanks, this was one of the most helpful posts I've seen so far. In my question I have a map $p: A' \rightarrow A$. Does your set up reverse the roles of $A$ and $A'$? $\endgroup$ – zahl Jun 19 '17 at 15:43
  • $\begingroup$ @zahl Glad it helped! Yes, in the diagram above, we were interested in showing natruality of $\eta$, and so naturally we consider a morphism $A\to A'$. This morphism induces $q:F(A)\to F(A')$, while it also gives a $p:A\to A'$, so we have reversed role of $A$ and $A'$ in definition of $p$ in adjunction (or we just set $A=A'$ and $A'=A$ respectively in definition of $p$). $\endgroup$ – user160738 Jun 20 '17 at 1:57

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