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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. enter image description here

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.

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  • $\begingroup$ 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 $\endgroup$
    – user8268
    Commented May 28, 2023 at 16:19
  • $\begingroup$ @user8268 thanks you ! I guess there's a mistake in the book then. $\endgroup$
    – Julia
    Commented May 28, 2023 at 16:25

1 Answer 1

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The author probably tries to mean a natural isomorphism.

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 $$Nat(F_1,F_2) \xrightarrow{\sim} Nat(G_2,G_1)$$

Natural transformations corresponding to one another are called mates under the adjunction. The interesting thing is that this result holds in a 2-category for adjoint pairs of 1-cells. This was first observed by G.M. Kelly in his paper, "Elements of 2-categories"

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