2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.