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

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

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

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

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.

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

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.

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