For a monad $(T,\mu,\eta)$ if $T(A) = T(B)$, does this imply that $\mu\_A = \mu\_B$? I want to know because in the bijection between Kleisli triples and monads, given a monad, we define $f^\* := T(f) ; \mu\_B$ if $f : A\to T(B)$ (c.f. [Prop 1.6](http://www.google.com/url?sa=t&source=web&ct=res&cd=1&ved=0CBgQFjAA&url=http%3A%2F%2Fciteseerx.ist.psu.edu%2Fviewdoc%2Fdownload%3Fdoi%3D10.1.1.79.733%26rep%3Drep1%26type%3Dpdf&rct=j&q=notions+of+computation+and+monads&ei=7Br9S9iXK-WT4gah5sihCw&usg=AFQjCNH28ATtI90zSFufGd-GBXGJQu7Drw)), but this needs to be well defined even when $T$ is not an embedding.


Clarification (more formal): Given a monad $(T,\mu,\eta)$, I need to define ${}^\*$ on the class $\{f \;|\; \exists A,B \in |\mathbf{C}| . f : A \to T B \}$. If $T$ were an embedding then I could just take $B := T^{-1}(\mathrm{cod}(f))$. What do I do though when $T$ is not an embedding? All I know about the codomain of an element of this class is that it is in the image of $T$. There may be many $B$'s and the $\mu\_B$'s may be different! 



Kleisli Monad bijection
In fact, the Kleisli star is a partial map ${}^\* : \mathrm{Mor}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \times \mathrm{Obj}(\mathbf{C}) \to \mathrm{Mor}(\mathbf{C})$. So there is no problem!


