Begin with an equivalence $F: C \to D$, $G: D \to C$ along with natural isomorphisms_ to the identities, so $\eta: 1_C \simeq GF$ and $\epsilon: FG \simeq 1_D$. The claim is that we can replace $\epsilon$ by $\epsilon'$ such that the resulting ($\eta$, $\epsilon'$) pair obey the triangle identities.

Begin by defining $\gamma \equiv G \overset{\eta G}{\rightarrow} GFG \overset {G \epsilon}{\rightarrow} G$. This need not be identity, and will not be if $(\eta, \epsilon)$ is not an adjunction. So define $\epsilon' \equiv \epsilon \circ F\gamma^{-1}$. The claim is that by "the naturality of the $\eta$ the diagram below commutes, which proves one triangle identity $G \epsilon' \circ \eta G = 1_G$ ":

enter image description here

I do not understand many things about this diagram:

  1. Which part of the diagram is $G \epsilon'$?
  2. Which part of the diagram is $\eta G$?
  3. Why does the left triangle commute?
  4. Does the right triangle commute because of the definition of $\gamma$?
  5. Does the commutativity of the whole diagram follow from the commutativity of the left and right triangles?

My answers to the questions are these, which I feel are probably wrong:

  1. $G\epsilon' = G\epsilon \circ GF \gamma^{-1}$. So the right-hand side of the top row is $G \epsilon'$
  2. $\eta G$ is the left-hand side of the top row.
  3. Computing the bottom leg, we get: $\eta G \circ \gamma^{-1} = \eta G \circ (G \epsilon \circ \eta G)^{-1} = \eta G \circ (\eta G)^{-1} \circ (G \epsilon)^{-1} = (G\epsilon)^{-1}$. Computing the top leg, we get $GF\gamma^{-1} \circ \eta G = GF((G\epsilon)^{-1} \circ (\eta G)^{-1}) \circ \eta G$. I'm stuck here, I'm not sure how to reduce this.
  4. By definition, $\gamma = G \epsilon \circ \eta G$, from which it follows that the right triangle commutes.
Full Snippet, Prosition 4.4.5, Promoting equivalence to adjoint equivalence

enter image description here


1 Answer 1

  1. The second and third horizontal natural transformations are $G\epsilon \circ GF \gamma^{-1} = G(\epsilon \circ F\gamma^{-1}) = G \epsilon'$.
  2. I don't quite follow this question - there are two arrows labelled $\eta G$.
  3. That's naturality of $\eta G : G \implies GFG$ applied to $\gamma^{-1}$.
  4. Yes.
  5. Yes, this is a general fact, it is not specific to this situation.

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