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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.

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    $\begingroup$ What is $\varepsilon_{\alpha}$? $\endgroup$ – Oskar Nov 12 '18 at 0:25
  • $\begingroup$ @Oskar Sorry. It's a typo. I mean $\mathcal{E}_{\alpha}$ (the functor $\mathcal{E}$) $\endgroup$ – Carlos Jiménez Nov 12 '18 at 0:30
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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.

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  • $\begingroup$ This is a very nice proof. +1 $\endgroup$ – Matematleta Nov 14 '18 at 3:14

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