I'm stuck with the next exercise. I don't know how can I solve it.

Let $\mathcal{F}:\mathcal{C}\to\mathcal{D}$ and $\mathcal{E},\mathcal{G}:\mathcal{D}\to\mathcal{C}$ three adjoint functors $\mathcal{E}\dashv\mathcal{F}\dashv\mathcal{G}$. Let $(\alpha,\beta)$ the unit and counit of the adjunction $\mathcal{E}\dashv\mathcal{F}$ and $(\eta,\varepsilon)$ the unit and counit of the adjunction $\mathcal{F}\dashv \mathcal{G}$. Suposse that the next conditions (1) holds:

a) $\text{Id}_{\mathcal{EFE}}=\mathcal{E}_{\alpha}\circ\beta_{\varepsilon}$

b) $\text{Id}_{\mathcal{FEF}}=\alpha_{\mathcal{F}}\circ\mathcal{F}_{\beta}$

Prove that the next conditions holds

a) $\text{Id}_{\mathcal{FGF}}=\mathcal{F}_{\eta}\circ\varepsilon_{\mathcal{F}}$

b) $\text{Id}_{\mathcal{GFG}}=\eta_{\mathcal{G}}\circ \mathcal{G}_{\varepsilon}$

First, I proved that the condition a) is equivalent to the condition b), thus, I only need to prove one of a) or b). Next, I wrote the commutative diagrams related to my hypothesis:

By (1) we have the next commutative diagrams

enter image description here

And because $\mathcal{E}\dashv\mathcal{F}$ and $\mathcal{F}\dashv\mathcal{G}$ we have the next pair of commutative diagrams

enter image description here

enter image description here

Thus I need to prove that the next diagram is commutative:

enter image description here

But I don't know how. By combining the previous diagrams, I obtained the next commutative diagram:

enter image description here

But in fact that diagram does not help. I saw that if I prove the previous diagram, then in fact I will prove that $\mathcal{F}_{\eta}$ is a natural isomorphism, but, again, I'm stuck. Any hint? I really appreciate any help you provide me.

  • 1
    $\begingroup$ What is $\varepsilon_{\alpha}$? $\endgroup$
    – Oskar
    Nov 12, 2018 at 0:25
  • $\begingroup$ @Oskar Sorry. It's a typo. I mean $\mathcal{E}_{\alpha}$ (the functor $\mathcal{E}$) $\endgroup$ Nov 12, 2018 at 0:30

1 Answer 1


This is one possible way to approach the problem.

By the definition of adjunction we have the following chain of natural isomorphisms $$\newcommand{\mc}{\mathcal} \newcommand{\id}{\text{Id}} \begin{align*} \mc D[\mc{FGF}(c),\mc{FGF}(c)] &\cong \mc C[\mc{EFGF}(c),\mc{GF}(c)] \\ &\cong \mc D[\mc{FEFGF}(c),\mc{F}(c)] \\ &\cong \mc C[\mc{EFEFGF}(c),c] \end{align*}\ . $$ Since these are natural bijections we have that $$\mc F(\eta)\circ \epsilon_{\mc F}=\id_{\mc{FGF}}$$ holds only and only if their transformed morphisms in $\mc C[\mc{EFEFGF}(c),c]$.

From there an additional small step is required to complete the solution. Here it follows this small step but I suggest you take a look at it only after having tried by yourself.

In the same way you can prove the other identity.

I hope this helps.

A computation shows that $\mc F(\eta_c) \circ \epsilon_{\mc F(c)}$ becomes $\beta_c \circ \mc{E}(\epsilon_{\mc F(c)})\circ \beta_{\mc{EFGF}(c)}$ while $\id_{\mc{FGF}(c)}$ becomes $\beta_c \circ \mc E(\epsilon_{\mc F(c)})\circ \mc{EF}(\beta_{\mc{GF}(c)})$. To conclude we can just observe that $$ \begin{align*} \mc{EF}(\beta_{\mc{GF}(c)}) &= \mc{EF}(\beta_{\mc{GF}(c)})\circ\mc E(\alpha_{\mc{FGF}(c)})\circ\beta_{\mc {EFGF}(c)}\\ &= \mc{E}(\mc F(\beta_{\mc{GF}(c)})\circ \alpha_{\mc{FGF}(c)})\circ \beta_{\mc{EFGF}(c)} \\ &= \beta_{\mc{EFGF}(c)} \end{align*}$$ where the first equality follows from the hypothesis, the second by functoriality and the third by the triangle identities for adjunctions.

  • $\begingroup$ This is a very nice proof. +1 $\endgroup$ Nov 14, 2018 at 3:14

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.