CoKleisli category of the induced comonad of a monad Given a monad, it induces a comonad on its Eilenberg-Moore category. We can then take the coKleisli category of this comonad. Can we say anything interesting about this?
 A: Let $\mathcal{C}$ be a category, let $\mathbb{T}$ be a monad on $\mathcal{C}$, let $\mathcal{D} = \mathcal{C}^\mathbb{T}$ be the category of $\mathbb{T}$-algebras, let $\mathbb{G}$ be the induced comonad on $\mathcal{D}$, and let $\mathcal{E}$ be the Kleisli category associated with $\mathbb{G}$. By the universal property of the Kleisli category, we have the following commutative diagram,
$$\require{AMScd}
\begin{CD}
\mathcal{D} @= \mathcal{D} \\
@V{H}VV @VV{U}V \\
\mathcal{E} @>>{R}> \mathcal{C}
\end{CD}$$
where $H : \mathcal{D} \to \mathcal{E}$ is the cofree $\mathbb{G}$-coalgebra functor and let $U : \mathcal{D} \to \mathcal{C}$ is the forgetful functor.
The functor $R : \mathcal{E} \to \mathcal{C}$ is fully faithful. (This is a general fact about comonads and their Kleisli categories.) Indeed, writing $V : \mathcal{E} \to \mathcal{D}$ for the forgetful functor and $F : \mathcal{C} \to \mathcal{D}$ for the free $\mathbb{T}$-algebra functor, we have
$$\mathcal{E} (H A, H B) \cong \mathcal{D} (V H A, B) = \mathcal{D} (F U A, B) \cong \mathcal{C} (U A, U B)$$
which is essentially how the functor $R : \mathcal{E} \to \mathcal{C}$ is defined in the first place. So we deduce that $\mathcal{E}$ is equivalent to the full subcategory of $\mathcal{C}$ spanned by the image of $U : \mathcal{D} \to \mathcal{C}$.
Things are more interesting if you instead look at the category of $\mathbb{G}$-coalgebras. In that case, we get a commutative diagram of the form below,
$$\begin{CD}
\mathcal{C} @>{L}>> \mathcal{D}_\mathbb{G} \\
@V{F}VV @VVV \\
\mathcal{D} @= \mathcal{D}
\end{CD}$$
where the right vertical arrow is the forgetful functor. The functor $L : \mathcal{C} \to \mathcal{D}_\mathbb{G}$ does not necessarily have any good properties. If $L$ is fully faithful, then we say $\mathbb{T}$ is of descent type; and if $L$ is an equivalence of categories, then we say $\mathbb{T}$ is of effective descent type. For an example of this, look at the case where $\mathcal{C}$ is the category of (left) $R$-modules for some ring $R$ and $\mathbb{T}$ is the monad for (left) $S$-modules for some faithfully flat $R$-algebra $S$.
A: Let $T:{\bf A}\to{\bf A}$ be the monad with $\mu:T^2\to T$. Let $U:{\bf A}^T\to{\bf A}$ be the forgetful functor from the Eilinberg-Moore category, mapping $(A,\alpha)\mapsto A$, and let $\tilde T:{\bf A}\to{\bf A}^T$ be its left adjoint, mapping $A\mapsto (TA,\,\mu_A)\ $ (which plays the role of the 'free $T$-algebra' over $A$).
Then the comonad belonging to this adjunction is $S:=\tilde T\circ U$, and by definition, its co-Kleisli category $({\bf A}^T)_S$ consists of objects of ${\bf A}^T$ as objects and for $T$-algebras $\mathcal A:=(A,\alpha)$ and $\mathcal B:=(B,\beta)$, we have
$$\hom_{({\bf A}^T)_S}(\mathcal A,\,\mathcal B):=\hom_{{\bf A}^T}(S\mathcal A,\,\mathcal B)\,.$$
But, if we apply $S=\tilde T\circ U$ and use the adjunction $\tilde T\dashv U$, we get
$$\hom_{({\bf A}^T)_S}(\mathcal A,\,\mathcal B)=\hom_{{\bf A}^T}(S\mathcal A,\,\mathcal B)=\hom_{{\bf A}^T}(\tilde T A,\,\mathcal B)\cong \hom_{\bf A}(A,\,U\mathcal B)=\hom_{\bf A}(A,B)\,.$$
You can verify that also composition of arrows in $({\bf A}^T)_S\,$ correspond to composition in $\bf A$.
Also observe that $\mathcal A\cong\mathcal B$ here iff $A\cong B$ in $\bf A$.
Hence, the co-Kleisli category is equivalent to the full subcategory of $\bf A$ spanned by those objects $A$ which admit any $T$-algebra structure.
