TITLE: Monad = Reflective Subcategory?
QUESTION [9 upvotes]: After answering this question here about Kleisli triples, I realized that this whole Kleisli triple construction: 

$T:{\rm Ob}\mathcal C\to{\rm Ob}\mathcal C$, $\ \eta_A:A\to TA$ for all $A\in {\rm Ob}\mathcal C$ and
  $f\mapsto f^* $ for all $f:A\to B^T$ such that

$\eta_A^*=1_{TA}$
$\eta_Af^*=f\ $ for all $\ f:A\to TB\ $  (writing composition from left to right)
$(fg^*)^*=f^*g^*\ $ for all arrows $\ f:A\to TB,\ g:B\to TC$.


determines nothing else but a reflective subcategory $\ \tilde{\mathcal C}\ $ of $\ \mathcal C$ (consisting of exactly the arrows of the form $f^*$, among objects of the form $TA$):
The reflection of any $\ A\in{\rm Ob}\mathcal C\ $ to $\ \tilde{\mathcal C}$ is given by $\eta_A$, and then conditions  2. states exactly the reflection property (any $f$ from $A$ to $\tilde{\mathcal C}$, that is, any $f:A\to TB$ uniquely factors through $\eta_A$), and 1. and 3. ensure that $\tilde{\mathcal C}$ will be a subcategory.
Conversely, if a reflective subcategory $\tilde{\mathcal C}$ is given, we can fix a reflection arrow to $\tilde{\mathcal C}$ from each object $A\in{\rm Ob}\mathcal C$, and call them $\eta_A$, and its codomain $TA$. Then, as $TB\in{\rm Ob}\tilde{\mathcal C}$, for each $f:A\to TB$ we have a unique factorisation through $\eta_A$, and this determines $f^*$ so that $f=\eta_Af^*$.

Questions: 

Is it a well know fact that monads thus are basically the same as reflective subcategories (at least up to natural isomorphism)?
Is this argument deficient in any point?

Update:
Question 3. Starting out from a (not full) reflective subcategory $\mathcal B$ of $\mathcal C$, then constructing $\tilde{\mathcal C}$ by arbitrarily fixed reflection arrows $\eta_A$ as above, I can see that $\tilde{\mathcal C}\subseteq\mathcal B$ is a full reflective subcategory. Are they necessarily equivalent? If not, what more can we say about them?

REPLY [6 votes]: I think I have found it.
So, the situation is indeed that each (not necessarily full) reflective subcategory determines a monad, and each monad determines a Kleisli triple that determines a reflective subcategory.
But, these procedures are not inverses to each other in both directions (they rather establish an adjoint situation only).

If we start out from a monad $T:\mathcal C\to\mathcal C$, we construct $\tilde{\mathcal C}$, then consider its monad $\tilde T$, we get the same:
$T\simeq \tilde T$ (naturally isomorphic).
On the other hand, if we start out from a reflective subcategory $\mathcal B$ of  $\mathcal C$, and fix reflections for each object $C\in{\rm OB}\mathcal C$ (that is, pick a left adjoint of the inclusion $\mathcal B\hookrightarrow\mathcal C$), and construct $\tilde{\mathcal C}$,
we can well have that $\tilde{\mathcal C}$ is much smaller than (and not equivalent to) $\mathcal B$.

For an example, consider an arbitrary embedding $H:\mathcal{Grp}\hookrightarrow\mathcal{Set}$ (for example $H(G):=\{G\}\times UG$ where $UG$ is the underlying set of $G$), and let $\mathcal B$ be the image of $\mathcal{Grp}$ in $\mathcal{Set}$, consisting of the 'group homomorphisms' between these underlying sets.
Then, $\mathcal B$ is a reflective subcategory of $\mathcal{Set}$, and the reflection is the free group functor, so in this case $\widetilde{\mathcal{Set}}$ will consist only the ($H$-underlying sets of) free groups, which is strictly smaller than $\mathcal B$ itself.
However, in general, by the construction of $T$ for the reflective $\mathcal B\le\mathcal C$, we have 
$$\tilde{\mathcal C}\le \mathcal B\hookrightarrow \mathcal C^T$$
where $\mathcal C^T$ is the (Eilenberg-Moore) category of $T$-algebras, and the first containment is full and reflective.