Let $F\colon\mathsf{C}\to\mathsf{D}, G\colon\mathsf{D}\to\mathsf{C}$ together with $\eta\colon 1_{\mathsf{C}}\to \mathsf{GF}, \epsilon\colon\mathsf{FG}\to 1_{\mathsf{D}}$ be an equivalence of categories. I know how to replace $\epsilon$ with $\epsilon'$ and such $\epsilon'$ is still an natural isomorphism and together with $\eta$ they form a unit-counit pair.

Howerver, for some reason I can't figure out with what we should replace $\eta$ if we want that instead. Riehl in the book Category Theory in Context defines $\gamma = G(\epsilon)\circ \eta G$ and claims that

Redefining either $\epsilon$ or $\eta$ so as to absorb the isomorphism $\gamma$ - it will not matter which -we will show that there sulting pair of natural isomorphisms define the unit and the counit of an adjunction $F\dashv G$.

Then she replaces $\epsilon$ with $\epsilon' = \epsilon\circ F(\gamma^{-1})$ and proves that $\eta$ and $\epsilon'$ satisfy the triangle identites for the unit and the counit. I tried replacing $\gamma$ with $\epsilon F\circ F(\eta)$ to no avail.

  • $\begingroup$ Do you want to exchange the roles of $\epsilon$ and $\eta$, or of $F$ and $G$? $\endgroup$
    – Berci
    Sep 2, 2019 at 17:51
  • $\begingroup$ So want to replace $\eta$ by something else to obtain an adjunction ? $\endgroup$
    – Arnaud D.
    Sep 2, 2019 at 19:21
  • $\begingroup$ @Berci The roles of $\epsilon$ and $\eta$. $\endgroup$
    – Jxt921
    Sep 2, 2019 at 20:17
  • $\begingroup$ @ArnaudD. Exactly. $\endgroup$
    – Jxt921
    Sep 2, 2019 at 20:17

1 Answer 1


You don't need to replace $\gamma$.

Set $\ \eta':=\gamma^{-1}F\circ\eta$.

Then we have $G(\varepsilon)\circ \eta'G\ =\ G(\varepsilon)\circ \eta^{-1}GFG \circ G\varepsilon^{-1}FG\circ\eta G$. $$\matrix{GFGx &\overset{\eta GFG}\to& GFGFGx \\ {\scriptstyle G\varepsilon}\downarrow \phantom{ {}_{G\varepsilon}} && \phantom{{}_{G\varepsilon FG}} \downarrow {\scriptstyle G\varepsilon FG}\\ Gx &\underset{\eta G}\to& GFGx}$$ Show that this square is commutative, and as each arrow is invertible, the above composition is $1_{Gx}$.

Similarly, $\varepsilon F\circ F(\eta')\ =\ \varepsilon F\circ F\eta^{-1}GF\circ FG\varepsilon^{-1}F\circ F\eta\ =\ 1_{Fx} $, belonging to the commutativity of $$\matrix{FGFx &\overset{F \eta GF}\to& FGFGFx \\ {\scriptstyle \varepsilon F}\downarrow \phantom{ {}_{\varepsilon F}} && \phantom{{}_{FG\varepsilon F}} \downarrow {\scriptstyle FG\varepsilon F}\\ Fx &\underset{F\eta}\to& FGFx}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.