why T-alg is terminal in T-adj

Question

In the following $(\mathscr D,T, \eta,\mu)$ is a monad

QUESTION: Why is K unique?

Let $K’$ be another such. Since $G^T$ is faithful it suffices to prove $K=K’$ on objects. But $G^T$ isn’t injective on objects and F isn’t surjective on objects…

Answer 1

Suppose $K$ has this property, and fix $c$ an object of $\mathcal{C}$. Then the underlying object of $K c$ is $G^T(K c) = G(c)$. Moreover, if we let $\varepsilon$ be the counit of the adjunction between $F$ and $G$, then we know that $K(\varepsilon_c)$ is a morphism of $T$-algebras. However, the domain of $K(\varepsilon_c)$ is $K(FG(c)) = F^T(Gc) = (T(Gc), \mu_{Gc})$; and the comain is $K(c)$. I will leave it as an exercise to show that more generally, if $\beta : (TX, \mu_X) \to (X, \alpha)$ is a morphism of $T$-algebras, and if $\beta \circ \eta_X = \operatorname{id}_X$, then $\alpha = \beta$. In our case, we have $G^T (K \varepsilon_c) \circ \eta = G \varepsilon_c \circ \eta = \operatorname{id}$ since by assumption, the unit of the adjunction between $F$ and $G$ gives rise to the unit of the monad $T$. Thus, we conclude that $Kc$ as a $T$-algebra must be equal to $(Gc, G^T(K \varepsilon_c)) = (Gc, G \varepsilon_c)$. Since, as you observed, $G^T$ is faithful, this suffices to show that $K$ is uniquely determined.

Answer 2

$G^TK=G$ tells you that $K$ should take an object $c$ to a $T-$algebra structure on $Gc$.

The second fact is that the $T$ algebra structure ($Td\to d $) on an object of $T-Alg$ is given by the (bottom arrow in the commuting square )of the component of the counit of the adjunction between $D,D^T$(this is just checking the definitions ).

The third fact is that $K$ commutes with the counits of the adjunctions owing to the fact they induce the same monad on $D$(this is a diagram chase). For the above reasons the action of $K$ on objects is forced. (Note that,the action on morphisms is straightforward, you need that as well($G^TK=G$)).