Regarding a difficulty in the Fakir article about associated idempotent triple I just had post this question in SE: https://math.stackexchange.com/questions/518054/about-details-of-the-fakir-theorem-proof-associated-idempotent-triple but dont get any answer.
I understand that at the first sight it seems a silly/trivial verifications (to me too) but I discussed the matter with some known person (PhD) and my doubt seems to be well founded.
On the ncatlab work http://ncatlab.org/toddtrimble/published/Associated+idempotent+monad+of+a+monad  Todd Trimbe quote the Fakir theorem in [1] about the associated idempotent triple:
Let $(T, \eta  , \mu)$ a  triple on a complete category  $\mathscr{C}$
in his article [1] Fakir claim to define a triple $(T', \eta', \mu')$ as follow: let  $T'$ the Kernel:   $T' \xrightarrow{k_X} T \rightrightarrows T \circ T$ where the couple is given by $\eta  T$ and $T\eta $. From $\eta T \ast \eta  = T\eta \ast \eta $ (apply  $\eta$ to $\eta_X$)  follow $\eta': 1 \Rightarrow  T'$ with $\eta_X= k_X\circ  \eta'_X $. We observe that:
1)  $\mu_X \circ  T(\eta_X)= 1_{T(X)}$  and then $\mu_X\circ T(k_X)\circ T(\eta'_X)=1$ .
For obtain $\mu': T'T' \Rightarrow  T'$ we consider that  $T'T'(X)$ is defined as the follow Kernel: 
$T'(T'(X)) \xrightarrow{k_{T'(X)}} T(T'(X)) \rightrightarrows (T \circ T)(T'(X))$ where the couple is given by
$\eta_{TT'X}$ and $T(\eta_{T'X}) $.
In the article [1] Fakir claim to obtain a morphism $\mu': T'T'(X) \to  T(X)$ from the universal property of kernel, assuming (implicitly) that $\mu_X\circ T(k_X)\circ k_{T'X}$ equalize the couple  $\eta_{TX},\ T\eta_X: T(X) \to T T(X)$, observe that from (1)  $\mu_X\circ T(k_X)$ cannot equalize this couple.
Then I consider the diagram 
$$\begin{array}{ccccc}
T'T'X & \xrightarrow{k_{T'X}} & TT'X & \xrightarrow[T\eta_{T'X}]{\eta_{TT'X}} & TTT'X \\
&& T(k_X)\downarrow && \downarrow TT(k_X)\\
& & TTX & \xrightarrow[T\eta_{TX}]{\eta_{TTX}} & TTTX \\
&& \mu_X\downarrow && \downarrow f\\
& & TX & \xrightarrow[T\eta_X]{\eta_{TX}} & TTX \\
\end{array}$$
If this is (mutually) commutative we are done. The top square is mutually commutative, but the below isn't if we put $f=T\mu$ or $f=\mu_T$.
How can we prove the existence of $\mu'$?.
I wish a proof in term of natural transformation, seems that in terms of string diagrams the things work well, but I dont know how translate in "classical" terms. 
Biblio:
[1] Fakir, Monade idempotente associee a une monade, C. R. Acad. Sci. Paris Ser. A 270 (1970), 99-101
 A: Yes, we first want to show that the composites of those squares are serially commutative, taking $f = T\mu$. In other words (let me drop the $X$; it plays no role) that 


*

*$\eta T \circ \mu \circ Tk = T\mu \circ TTk \circ \eta TT'$, and 

*$T\eta \circ \mu \circ Tk = T\mu \circ TTk \circ T\eta T'$. 
Now the first of these is trivial because it's just an instance of a naturality square for $\eta$. For the second, notice that both sides of the asserted equation are $T$-algebra maps. To check the equality of $T$-algebra maps $\phi, \psi$ when their common domain is a free $T$-algebra $T T'$, it suffices to show that $\phi \circ \eta T' = \psi \circ \eta T'$ (this is the universal property of free algebras). 
So we have to check that 


*

*$T\eta \circ \mu \circ Tk \circ \eta T' = T\mu \circ TTk \circ T\eta T' \circ \eta T'$. 


The left side is $T\eta \circ \mu \circ \eta T \circ k$ (by $\eta$-naturality), and reduces to $T\eta \circ k$ since $\mu \circ \eta T$ is an identity morphism. 
The right side is $T\mu \circ TTk \circ \eta TT' \circ \eta T'$ (by $\eta$-naturality), and we apply $\eta$-naturality a few more times: 
$$\begin{array}
T\mu \circ TTk \circ \eta TT' \circ \eta T' & = & T\mu \circ \eta TT \circ Tk \circ \eta T' \\ 
 & = & T\mu \circ \eta TT \circ \eta T \circ k \\ 
 & = & \eta T \circ \mu \circ \eta T \circ k
\end{array}$$ 
and this reduces to $\eta T \circ k$ since $\mu \circ \eta T$ is an identity. 
So after the reductions, it boils down to the equality $T\eta \circ k = \eta T \circ k$, which is true since $k$ is the equalizer of $T\eta, \eta T$. 

It should be obvious to Buschi Sergio that this is enough, but for anyone else reading out there: the required map $\mu': T'T' \to T'$ is defined to be the unique morphism such that $\mu \circ Tk \circ kT' = k \circ \mu'$, where the existence of $\mu'$ would follow (since $k$ is the equalizer of $T\eta, \eta T$) from the fact that 
$$T\eta \circ \mu \circ Tk \circ kT' = \eta T \circ \mu \circ Tk \circ kT'.$$ 
But the serial commutativity shown above allows us to rewrite this asserted equation as 
$$T\mu \circ TTk \circ T\eta T' \circ kT' = T\mu \circ TTk \circ \eta TT' \circ kT'$$ 
which clearly holds since $kT'$ equalizes $T\eta T', \eta TT'$. 
