How do I show that, if $fp=fq$ then $p=q$, in the context of $f$ being the counit of an adjunction and $p,q$ natural transformations? $\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}\newcommand{\I}{\mathscr{I}}\require{AMScd}$The titular statement is probably false, but it was the only way I could see to fit my question into the character limit. This is from an exercise in Tom Leinster's Basic Category Theory.

Given a functor $F:\A\to\B$ and a category $\I$, there is a functor $F^\ast:[\B,\I]\to[\A,\I]$ defined on objects $Y\in[\B,\I]$ by $Y\mapsto Y\circ F$ and on arrows $\alpha$ by $\alpha\mapsto\alpha F$.
Given categories $\A,\B,\I$ and any adjunction (if anyone knows how to format this better, I'd be much appreciative! I can't get the arrows going both ways on the same row): $$\begin{CD}\A@>F>>\B\\@.\perp\\\A@<<G<\B\end{CD}$$Show that this gives rise to an adjunction: $$\begin{CD}[\A,\I]@>G^\ast>>[\B,\I]\\@.\perp\\ [\A,\I]@<<F^\ast<[\B,\I]\end{CD}$$

And I am basically nearly there. The way I personally prefer to think about adjoints is via the specification of terminal/initial morphisms, since that uniquely specifies an adjunction with minimal structure. Let $\eta,\varepsilon$ be the unit and counit of $F\dashv G$; all I need to do is show the existence of a terminal morphism $(F^\ast(X),\varepsilon_X^\ast)$ from $G^\ast$ to $X$ for each functor $X:\B\to\I$.
From now on, $B,B',p$ will refer to any arbitrary arrow $B\overset{p}{\to}B'$ in $\B$. To explore this, I first decide that, with $G^\ast F^\ast X=XFG$, the completion of this diagram obviously has a natural choice $\varepsilon^\ast_{X,B}:=X(\varepsilon_B)$: $$\begin{CD}XFG(B)@>XFG(p)>>XFG(B)\\@V\varepsilon^\ast_{X,B}VV@VV\varepsilon^\ast_{X,B'}V\\X(B)@>>X(p)>X(B')\end{CD}$$And it is not hard to see that that choice of $\varepsilon$ will make the square commute. The assertion of terminality means that if $Y\in[\A,\I]$ and $f:G^\ast Y\to X$ is any natural transformation, then there is one and only one natural transformation $\alpha:Y\to F^\ast X$ so that $f=\varepsilon_XG^\ast\alpha$. Consulting the squares, this means that: $$\begin{align}\tag{1}X(p)\varepsilon^\ast_{X,B}\alpha_{G(B)}&=X(p)f_B\\&\,\,\|\\\varepsilon^\ast_{X,B’}\alpha_{G(B')}YG(p)&=f_{B'}YG(p)\end{align}$$My idea to attack this: if $f$ naturally transforms $G^\ast Y\to X$ then in particular the following diagram commutes for all $B\in\B$: $$\begin{CD}YGFG(B)@>YG(\varepsilon_B)>>YG(B)\\@Vf_{FG(B)}VV@VVf_BV\\XFG(B)@>>X(\varepsilon_B)=\varepsilon^\ast_{X,B}>X(B)\end{CD}$$So if $(1)$ holds then I can right-compose both sides with $YG(\varepsilon_B)$ and obtain, via the triangle identities: $$\begin{align}\tag{2}X(p\varepsilon_B)\alpha_{G(B)}YG(\varepsilon_B)&=X(p\varepsilon_B)f_{FG(B)}\\\implies X(p\varepsilon_B)\alpha_{G(B)}&=X(p\varepsilon_B)f_{FG(B)}Y(\eta_{G(B)})\end{align}$$Using the commutative diagram above, it is not too hard to see that if we construct $\alpha_A:=f_{F(A)}\circ Y(\eta_A)$, we have a solution to the terminal morphism problem. However, a terminal morphism needs this $\alpha$ to be unique: I would love to be able to simply omit the $X(p\varepsilon_B)$ term on both sides of $(2)$ and reduce from $G(B)$ to any $A\in\A$, because then we would have a proof that $(1)$ implies $(2)$ implies $\alpha_A=f_{F(A)}Y(\eta_A)$, but I can't do that.
Or can I? My question to this forum is: how do I get rid of $X(p\varepsilon_B)$ on both sides of $(2)$ (if I even need to at all?) and how can I justify: equality over all $G(B)$ implies equality over all $A$? Maybe the definition of equality of natural transformations requires that the intermediary arrows be equal, so I can perhaps drop the $p$ - but I am still left with $X(\epsilon_B)$ which needs ridding of.
 A: I want to suggest an alternative solution to you, since your notation is rather involved. Given the adjunction $F\dashv G$ with its unit $\eta:1\rightarrow GF$ and counit $\varepsilon: FG \rightarrow 1$, you like to show that $F^* \vdash G^*$ is an adjunction with unit $\eta^*:1\rightarrow F^*G^*$ and counit $\varepsilon : G^*F^*\rightarrow 1$. For this it is enough to check the triangle identities. I will write down a proof of one of them, the other one is similar.
To check that the triangle identity
\begin{align*}
 F^* \xrightarrow{\eta^*F^*} F^*G^*F^* \xrightarrow{F^*\varepsilon^*} F^* = F^*\xrightarrow{id}F^*
\end{align*}
holds, it is enough to evaluate it at an arbitrary functor $X$ and show that the resulting equation holds. So take some $X$. We get
\begin{align*}
 XF \xrightarrow{XF\eta} XFGF \xrightarrow{X\varepsilon F} XF = XF \xrightarrow{id}XF
\end{align*}
But the equation above is clearly true, because $\eta$ and $\varepsilon$ satisfy the triangle identity.
But why would we expect that the transformations $\varepsilon^*$ and $\eta^*$ satisfy the triangle identities? The reason is that the operation $(\quad)^*$ is a 2-endofunctor of the 2-category $\mathtt{Cat}$ of categories and transformations (it reverses the direction of 1-cells though). As such it maps equations between 2-cells to equations between 2-cells just an ordinary functor preserves equations between 1-cells. In consequence we directly see that the triangle identities for $\eta$ and $\varepsilon$ imply those of $\eta^*$ and $\varepsilon^*$.
Edit: I would like to draw 2-cells as \Rightarrow, but there doesn't seem to be an \xRightarrow. Of course you get a two functor $(\quad)^*$ for each choice of $\mathscr I$.
A: $\newcommand{\A}{\mathscr{A}}\newcommand{\B}{\mathscr{B}}\newcommand{\I}{\mathscr{I}}\require{AMScd}$After some more thought, I have a solution which I leave here for my own reference and the reference and the reference of others.
It remains to show that: $$\tag{$\dagger$}\forall B\in\B,\,X(\varepsilon_B)\alpha_{G(B)}=X(\varepsilon_B)f_{FG(B)}Y(\eta_{G(B)})$$Implies: $$\forall A\in\A,\,\alpha_A=f_{F(A)}Y(\eta_A)$$Let $A\in\A$ be arbitrary and consider the commutative diagram (which follows as $\alpha:Y\to XF$ is a natural transformation): $$\begin{CD}Y(A)@>Y(\eta_A)>>YGF(A)\\@V\alpha_AVV@VV\alpha_{GF(A)}V\\XF(A)@>>XF(\eta_A)>XFGF(A)\end{CD}$$Right-composing $(\dagger)$ by $Y(\eta_A)$, with $B:=F(A)$ we get (via the triangle identities): $$\begin{align}X(\varepsilon_{F(A)})X(F(\eta_A)))\alpha_A&=X(\varepsilon_{F(A)})f_{FGF(A)}Y(\eta_{GF(A)})Y(\eta_A)\\\implies\alpha_A&=X(\varepsilon_{F(A)})f_{FGF(A)}Y(\eta_{GF(A)})\circ Y(\eta_A)\end{align}$$Now consider the commutative diagram ($f:YG\to X$ is natural): $$\begin{CD}YGFGF(A)@>YG(\varepsilon_{F(A)})>>YGF(A)\\@Vf_{FGF(A)}VV@VVf_{F(A)}V\\XFGF(A)@>>X(\varepsilon_{F(A)})>XF(A)\end{CD}$$Applying this to the equation gives: $$\alpha_A=f_{F(A)}\circ \underset{\mathbf{1}_{YGF(A)}}{\underbrace{Y(G(\varepsilon_{F(A)})\eta_{GF(A)})}}\circ Y(\eta_A)\quad\blacksquare$$
