Why is this diagram commutative? Adjunction of functors

Question

So I'm reading about adjoint functors and I've seen that if $F$ is left adjoint to $G$ and $\epsilon$ is the counit of the adjunction then the following diagram is commutative:

Where $\phi$ is the natural isomorphism yielding the adjunction and $\epsilon_A$ is the counit.

Could anyone explain why this diagram commutes?

Thanks very much!

Answer 1

$\require{AMScd}$

For $f:A \to B$ in $\mathbf{A}$, consider the diagram $$\begin{CD} \mathbf{A}(FGA,A) @>{\phi}>> \mathbf{X}(GA,GA)\\ @VVV @VVV \\ \mathbf{A}(FGA,B) @>{\phi}>> \mathbf{X}(GA,GB) \end{CD}$$ which commutes by naturality of $\phi$. Then take $\epsilon \in \mathbf{A}(FGA,A)$, it goes to $Id_{GA}$ on the right and then goes down to $Gf \in \mathbf{X}(GA,GB)$. In the other direction, $\epsilon$ goes down to $f\circ \epsilon \in \mathbf{A}(FGA,B)$ and then also goes to $Gf \in \mathbf{X}(GA,GB)$ by commutativity of the diagram.

This shows that your diagram commutes.

Answer 2

The bijection $\phi_{GA,B}$ is defined by $(f:FGA\to B)\mapsto (Gf\circ \eta_{GA}:GA\to GB)$. Applying this formula to $f=g\circ \epsilon_A$ yields$$\phi_{GA,B}(g\circ \epsilon_A)=G(g\circ \epsilon_A)\circ \eta_{GA}=G(g)\circ (G\epsilon_A\circ \eta_{GA})=G(g),$$because of the triangular identity $G\epsilon_A\circ \eta_{GA}=Id_{GA}$.