Does a Kleisli triple need naturality conditions? I'm reading a paper by Eugenio Moggi entitled "Notions of Computation and Monads". It introduces the concept of a “Kleisli triple” on a category $\mathcal C$, which is $(T, \eta, -^*)$, where:


*

*$T$ is an object map on $\mathcal C$,

*$\eta_A : A \to TA$,

*For $f : A \to TB$, $f^* : TA \to TB$;


satisfying laws:


*

*$\eta_A^* = 1_{TA}$

*$f^* \circ \eta_A = f$

*$f^* \circ g^* = (f^* \circ g)^*$


It then goes on to claim that these definitions give a monad. However, there doesn't seem to be anything requiring even $\eta$ to be a natural transformation, let alone the $\mu$ that you can define from the triple. Furthermore, when I searched around for more on Kleisli triples, this seems to be a common theme – no-one is concerned with naturality.
Are the naturality conditions all just implicit? Are they too obvious to write down? Or are they truly unnecessary?
 A: I must eat my words, it seems – although the first couple of articles I read ignored naturality, I've come across a set of notes that cover it.
The key is the observation that naturality conditions are defined in terms of the action of $T$ on morphisms, which is defined in terms of the other components of the triple as $Tf = (\eta_A \circ f)^*$. It's that which makes everything work out, although I think it's a bit of an omission of Moggi's not to at least acknowledge it.
As an example, here's the naturality of $\eta$: \[Tf\circ \eta_A = (\eta_B \circ f)^* \circ \eta_A = \eta_B \circ f\]
A: Here's another approach: (I write composition from left to right.)
Consider first the subcategory $\tilde{\mathcal C}\subseteq \mathcal C$ with objects $TA$ for all $A$ and with all arrows of the form $f^*$ (for $f:A\to TB$).
Then, consider the category $\mathcal T$ which consist of the disjoint union of $\mathcal C$ and $\tilde{\mathcal C}$, 
plus arrows from the left to the right, $A\dashrightarrow TX$ as original arrows 
of $\mathcal C$.  All compositions  are coming from $\mathcal C$.
(It is (the collage of) a profunctor $\mathcal C\not\to\tilde{\mathcal C}$.)
If $T$ was also given on the arrows, then,
for any $f:A\to B$  we should have a commutative diagram (also to ensure naturality of $\eta$)
$$\matrix{A &\overset{\eta_A}\longrightarrow &TA \\
\!\!\!f\downarrow &&\ \,\ \downarrow Tf \\
B &\underset{\eta_B}\longrightarrow &TB
} $$
The main thing is that $\eta_A$ will be a reflection of $A$  to $TA$ on the right side (that's why we had to take the subcategory $\tilde{\mathcal C}$), in particular will be epimorphism.
Now, the above diagram can define $Tf$ as
$$Tf:=(f\eta_B)^* $$
The universal property (i.e. reflections) guarantee that it will be indeed a functor. All the other details can be seen similarly. Note that for $h:A\to TB$, the $\eta_A h^*=h$ is heavily used.
For naturality of $\mu\ $ (defined as $\mu_A:=(1_{TA})^*:TTA\to TA$), you can apply that $\eta_A$ is epimorphic (in $\mathcal T$, that is, w.r.t. $\tilde{\mathcal C}$), and so
$$ \matrix{A &\overset{\alpha}\longrightarrow &TX \\
\!\!\!f\downarrow &\scriptstyle\#&\ \,\ \downarrow h^* \\
B &\underset{\beta}\longrightarrow &TY } 
\ \implies \quad 
\matrix{TA &\overset{\alpha^*}\longrightarrow &TX \\
\!\!\!\!\!Tf\downarrow &\scriptstyle\#&\ \,\ \downarrow h^* \\
TB &\underset{\beta^*}\longrightarrow &TY }
 $$
Now apply this to $X=A,\ Y=B,\ u:A\to B,\ h^*=Tu=f,$ and
$\alpha=1_{TA},\ \beta=1_{TB}$.
A: As you point out, a monad can be formed from a Kleisli triple where the naturality of $\eta$ and $\mu$ of the monad follow from the Kleisli laws and the identity $T f = (\eta_A \circ f)^\ast$, and also $\mu = \mathrm{id}_A^\ast$. We could equivalently view $(-)^\ast$ as a natural transformation $(-)^\ast_{A,B} \colon \mathcal{C}(A, T B) \rightarrow \mathcal{C}(T A, T B)$, where (for all $f \colon X \rightarrow A$, $g \colon B \rightarrow Y$):
$$
  \require{AMScd}
  \begin{CD}
    \mathcal{C}(A, T B) @>{\ast_{A,B}}>> \mathcal{C}(T A, T B) \\
    @V{\mathcal{C}(f, T g)}VV      @VV{\mathcal{C}(T f, T g)}V \\
    \mathcal{C}(X, T Y) @>>{\ast_{X,Y}}> \mathcal{C}(T X, T Y)
  \end{CD}
$$
That is, if $h \colon A \rightarrow T B$, then $T g \circ h^\ast \circ T f  = (T g \circ h \circ f)^\ast$. This again follows from the Kleisli laws and the definition of the morphism mapping $T f = (\eta_A \circ f)^\ast$.
