Recall that for a given adjunction $$F : \mathbb{C} \to \mathbb{D}$$ and $$G : \mathbb{D} \to \mathbb{C}$$ with $$F$$ left adjoint to $$G$$, the unit is given by $$\eta : 1_\mathbb{C} \Rightarrow GF$$ and the co-unit is given by $$\epsilon : FG \Rightarrow 1_\mathbb{D}$$. The adjunction induces an endofunctor on $$\mathbb{C}$$ given by $$T = GF$$. Moreover, we can define a 'unit' simply as $$\eta : 1_\mathbb{C} \Rightarrow GF$$ and a 'multiplication' $$\mu : T^2 \Rightarrow T$$ as $$\mu_C = G(\epsilon_{FC})$$. The claim is that this defines a monad structure on $$T$$. Let's verify that. We have to show that the following diagrams commute. Notice that all arrows of the diagrams live' in the functor category $$[\mathbb{C},\mathbb{C}]$$. I.e. they are natural transformations between endofunctors on $$\mathbb{C}$$.

First we verify the diagram for the unit.

For the left triangle, we have that $$G(\epsilon_{FC}) \circ \eta_{GFC} = id_{GFC}$$. For the right triangle, we have that $$G(\epsilon_{FC}) \circ GF\eta_C = G(\epsilon_{FC} \circ F(\eta_C)) = G(id_{FC}) = id_{GFC}$$.

The second diagram we need to check is the following. We need to verify that associativity of multiplication': $$G(\epsilon_{FC}) \circ GFG(\epsilon_{FC}) = G(\epsilon_{FC}) \circ G(\epsilon_{FGFC}).$$

However, here is where I'm stuck, does someone know the approach to verify this equation?

Use naturality of $$\varepsilon:FG\Rightarrow 1_{\Bbb D}$$ with the arrow $$\varepsilon_{FC}$$: $$\matrix{FG\,FGFC & \overset{\varepsilon_{FGFC}}\longrightarrow & FGFC \\ {FG\,\varepsilon_{FC}}\downarrow\phantom{FG\, \varepsilon_{FC}} && \phantom{\varepsilon_{FC}} \downarrow {\varepsilon_{FC}}\\ FGFC & \underset{\varepsilon_{FC}}\longrightarrow & FC}$$ and apply $$G$$ on this commutative square.
• You can right-click and see the latex code, alternatively you can initiate an edit to view it, then cancel. By the way, I use \matrix. – Berci Nov 3 at 13:40