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:enter image description here

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!

  • 1
    $\begingroup$ Non-answer: It’s category theory. Any sensible diagram commutes. $\endgroup$ – k.stm Oct 30 '15 at 20:56
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    $\begingroup$ Define "sensible." Prove this diagram is sensible :) $\endgroup$ – ebrahim Oct 30 '15 at 20:57


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.


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 $G\epsilon_A\circ \eta_{GA}=Id_{GA}$.


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