# Right adjoint unique up to isomorphism

i want to prove the following without the Yoneda Lemma (because it is the exercise): Suppose $F\dashv G$ (with unit $\eta$ and counit $\epsilon$) and $F\dashv G'$ (with unit \eta' and conunit $\epsilon'$) then $G\cong G'$.

I want to do this in three steps:

1. Construct $G(X)\rightarrow G'(X)$ and $G'(X)\rightarrow G(X)$ with help of the units and counits.
2. Show that this arrows are natural transformations.
3. Both arrows are each other inverse.

This is may plan so far. I will tell you what i have done so far and what my problems are:

• I used unit and counit to make $f:=G(X)\rightarrow G'(X)$ by $f_X=G'(\epsilon_X)\circ\eta_{G(X)}$. This is form my point of view a corollary of the triangle identity. On the same way we can define $g_X:=G'(X)\rightarrow G(X)$ by $g=G(\epsilon'_X)\circ\eta_{G'(X)}$.

My Question: Is this construction okay? If so then we can go to point 2, if not tell me why not and maybe you can help me further by correcting the following mistakes?

• Okay, second step. I have to show that if $h:X\rightarrow Y$, that then the following holds: $G'(h)\circ f_X=f_Y\circ G(h)$.

My problem: i do not see how to came form the one side of the equation to the other side. Is this trivally or what for computations must be done?

• I can't to the second step without help, so i can not do the third step. Can someone help me with this step, too?

I am very happy about each sort of help, information and solution. Thank you for reading and thinking about that. Thank you also for help.

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Why do you want to avoid the Yoneda Lemma, which provides a proof in one line? (math.stackexchange.com/questions/389933) – Martin Brandenburg Dec 19 '13 at 16:31

## 1 Answer

1. Yes, it seems ok.
2. Prove more generally that the constructions $F\alpha:=x\mapsto F(\alpha_x)$ and $\alpha G:=x\mapsto \alpha_{Gx}$ give natural transformations if $F,G$ functors and $\alpha$ is a natural transformation, and that composition of natural transformations ($\alpha\circ\beta:=x\mapsto \alpha_x\circ\beta_x$) is again natural.
So that, now we have the following composition of natural transformations: $$\varphi: G \ \overset{\eta'G}\longrightarrow\ G'FG \ \overset{G'\varepsilon}\longrightarrow\ G'$$ (what you called $f$).
3. We have $\varphi=G'\varepsilon\circ \eta'G$ and $\psi=G\varepsilon'\circ\eta G'$.

My favorit method is to draw squares for the units and counits with edges $1_{\Bbb A}$, $1_{\Bbb B}$ (omitted) and $F$, $G$ as below, and paste them together in appropriate ways. $$\matrix{&\overset{G'}\longrightarrow \\ &\ \ \ \ \ \ \ \varepsilon'\ \downarrow F \\ \ } \quad \quad \matrix{&& \\ F\downarrow\ \eta \\ \phantom{F\downarrow} \underset{G}\longrightarrow }$$ The squares are read from top right to bottom left, corresponding to e.g. $\varepsilon':FG'\to 1_{\Bbb B}$. Then consider the following pasting of squares: $$\matrix{\varphi & : &\varepsilon& \eta' & 1_{1_{\Bbb A}} \\ \psi & : &1_{1_{\Bbb B}} & \varepsilon' & \eta}$$

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