# Regarding a difficulty in the Fakir article about associated idempotent triple

I just had post this question here but in a less exact form .

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.

Let $(T, \eta , \mu)$ a triple on a complete category $\mathscr{C}$ he 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 $\mu'$?

Biblio:

[1] Fakir, Monade idempotente associee a une monade, C. R. Acad. Sci. Paris Ser. A 270 (1970), 99-101