Specific question about adjunction of functors

Question

Suppose to have three functors such that $F\dashv G, G':\mathbf D\to \bf C$. For every $c$ in $\mathrm{Ob}\bf C$, the adjunctions imply the existence of two universal arrows among the arrows of $F$ in $c$, call them $u:FGc\to c$ and $u':FG'c\to c$; so there must be two arrows, $f_c: Gc\to G'c$ and $f'_c: G'c\to Gc$, such that $u=u'\circ Ff$ and $u'=u\circ Ff'$. Then one proves that $f_c$ is an isomorphism with inverse $f_c'$, and also that the collection of $\{f_c\}_c$, as $c$ ranges in $\mathrm {Ob}\bf C$, constitute a natural isomorphism $G\xrightarrow\centerdot G'$.

However, always starting from $F\dashv G, G':\mathbf D\to \bf C$, one can observe that there are bijections natural in two variables, $\mathbf{ C}(-,G-)\cong \mathbf{ D}(F-,-)\cong \mathbf{ C}(-,G'-)$; from here, basically using Yoneda's lemma, one finds again a natural isomorphism from $G$ to $G'$. I can't see the link between these two natural isomorphisms $G\xrightarrow\centerdot G'$, despite probably being the same; thanks for any clarify.

Answer 1

The two natural transformations are indeed the same. I will also need the units of the adjunctions. I use standard notation: $\eta,\varepsilon$ are unit and counit of the adjunction $F\dashv G$ and $\eta',\varepsilon'$ are unit and counit of the adjunction $F\dashv G'$. My $\varepsilon$ is your $u$. As you have said, you have an isomorphism \begin{align} \mathcal D(d,Gc) \cong \mathcal C(Fd,c) \cong \mathcal D(d,G'c) \end{align} natural in both variables. By Yoneda's lemma there is exactly one morphism $g_c: Gc \to G'c$ which induces the above isomorphism, i.e. such that $\mathcal D(-, g_c)$ is the isomorphism $\mathcal D(-,Gc) \cong \mathcal D(-,Gc)$. You would like to show that the morphism $g_c$ you obtain this way is equal to $f_c$ which you have already defined. But what is the morphism $g_c$? Note that there is only one reasonable way to produce a morphism $Gc \to Gc'$ in such a general situation. You plug in $d = Gc$ in the above equation and map $1_{Gc}$ first to $\mathcal C(FGc,c)$ and then to $\mathcal D(Gc,Gc')$. It looks like this:

\begin{align} 1_{Gc} \mapsto \varepsilon_c \mapsto G'\varepsilon_c\, \circ\eta'_d\,. \end{align}

I claim that $g_c = G'\varepsilon_c \,\circ \eta_d'$. This is just a general fact which drops out of the proof of the Yoneda lemma, but you can check it if you like. To show that $f_c = g_c$ we need to verify that the diagram \begin{align}\require{AMScd} \begin{CD} FGc @>{\varepsilon_c}>> c\\ @V{Fg_c}VV @VV{1_c}V \\ FG'c @>{\varepsilon'_{c}}>> c \end{CD}\end{align} commutes, since $f_c$ is by definition the unique arrow which fits into it. But the above diagram commutes if and only if \begin{align}\require{AMScd} \begin{CD} Gc @>{G'\varepsilon_c\, \circ \,\eta_d'}>> G'c\\ @V{g_c}VV @VV{1_{G'c}}V \\ G'c @>{1_{G'c}}>> G'c \end{CD}\end{align} does commute by naturality of the adjunction $F\dashv G'$. The second diagram commutes by definition of $g_c$. There are some steps in the argument which might not be super clear if you are uncomfortable with adjunctions. Send me a comment if you don not understand a step, and I will edit the answer.

Answer 2

Let $\Phi$ be the natural isomorphism $\mathbf{D}(F-, -) \cong \mathbf{C}(-, G-)$ and $\Psi$ the natural isomorphism $\mathbf{D}(F-, -) \cong \mathbf{C}(-, G'-)$. From your description, I'm guessing that $u$ is $\Phi_{c, c}^{-1}(\mathrm{id}_{Gc})$ and $u'$ is $\Psi_{c, c}^{-1}(\mathrm{id}_{G'c})$. For brevity, I'll drop the subscripts and write $\Phi^{-1}(\mathrm{id}_{Gc})$ and $\Psi^{-1}(\mathrm{id}_{G'c})$. Then $f_c$ is $\Psi(\Phi^{-1}(\mathrm{id}_{Gc}))$ and $f'_c$ is $\Phi(\Psi^{-1}(\mathrm{id}_{G'c}))$.

The natural isomorphism $\mathbf{C}(-, G-) \cong \mathbf{D}(F-, -) \cong \mathbf{C}(-, G'-)$ in your second paragraph is $\Psi \circ \Phi^{-1}$. The natural isomorphism given by the Yoneda lemma (recall how the proof goes) is then $(\Psi \circ \Phi^{-1})(\mathrm{id}_{Gc}) = f_c$ with inverse given by $(\Psi \circ \Phi^{-1})^{-1}(\mathrm{id}_{G'c}) = f'_c$.