On the ncatlab work http://ncatlab.org/toddtrimble/published/Associated+idempotent+monad+of+a+monad Todd Trimbe quote the Fakir theorem about the associated idempotent triple, and this is based on the following construction:

Let $(T, \eta , \mu)$ a triple on a complete category $\mathscr{C}$, I wish define a triple $(T', \eta', \mu')$ where $T'$ 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 $\mu_X \circ T(\eta X)= 1_{T(X)}$ and then $\mu_X\circ T(k_X)\circ T(\eta'_X)=1$ .

Now for obtain $\mu': T'T' \Rightarrow T'$ we consider that $T'T'(X)$ is the 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} $.

Then to obtain $\mu'$we can find some morphism $T(T'X) \to T(X)$ that equalize the couple $\eta_{TX},\ T\eta_X: T(X) \to (T \circ T)(X)$.

I thought to $\mu_X \circ T(k_X): T(T'X) \to T(X)$, but from above this is a retraction, then a epimorphisms, then we would have that $\eta_{TX}= T\eta_X$ but this isn't true in general.

Then we need another chose for a morphism $T(T'X) \to T(X)$, but what?

Could someone explain this aspect of the Fakir proof more in details?

Sorry but I do not have access to the original work of Fakir:

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


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.