# A pair of adjoint functors and unicity

I was working on the following problem,

Given $$F_1,F_2:\mathscr{C}\rightarrow\mathscr{D}$$ and $$G_1,G_2:\mathscr{D}\rightarrow\mathscr{C}$$ such that $$(F_1,G_1)$$ and $$(F_2,G_2)$$ form adjoint pairs, show that if there is a morphism $$\varphi:G_1\rightarrow G_2$$ then there is a unique $$\varphi^*:F_2\rightarrow F_1$$ such that the following diagram commutes.

And show that if $$\varphi=1_{G_1}$$ then $$\varphi^*=1_{F_1}$$

I was able to do the first part of the exercice by using the Yoneda lemma to show that $$\begin{gather} \text{Nat}(G_1,G_2)\cong \text{Nat}(\mathscr{C}(-,G_1?),\mathscr{C}(-,G_2?))\cong \text{Nat}(\mathscr{D}(F_1-,?),\mathscr{D}(F_2-,?))\cong\text{Nat}(F_2,F_1) \end{gather}$$

However with $$\varphi =1_{G_1}$$ then $$G_1=G_2$$ hence $$F_1\cong F_2$$ being both right adjoint of the same functor.

But I am unable to understand why $$F_1=F_2$$ which I believe is necessary since I must show $$\varphi*=1_{F_1}$$.

There seem to be something obvious I am missing but I feel truly stuck.

• you are right, there is no reason for $F_1=F_2$; the only thing that you can show is that $\varphi^*$ is an isomorphism Commented May 28, 2023 at 16:19
• @user8268 thanks you ! I guess there's a mistake in the book then. Commented May 28, 2023 at 16:25

A side note : Given any two adjunctions $$F_1 \dashv G_1 : \mathcal{D} \to \mathcal{C}$$ and $$F_2, G_2 : \mathcal{D} \to \mathcal{C}$$, you see that there is a canonical bijection $$Nat(F_1,F_2) \xrightarrow{\sim} Nat(G_2,G_1)$$