Let $C$ be a symmetric monoidal category equipped with diagonals $\triangle_x: x \to x \otimes x$, that is, equipped with natural transformations $e_x: x \to 1$ and $\triangle_x : x \to x \otimes x $ such that $l_x \circ (e_x \otimes id_x) \circ \triangle_x = id_x$, where $l_x$ is the left unitor and the same holds for the right unitor.

Does it holds that if $(T,\eta, \mu)$ is a monoidal monad on $C$ (a monad object in the 2-category of lax monoidal endofunctors of $C$, more explicitly, see 1.2 [here][1]) then $Kl(T)$ is symmetric monoidal with diagonals? I couldn't find a proof for it and I'm a beginner with Kleisli categories so I will share my calculations.

Using that $T$ is a monoidal monad and that $C$ is symmetric I could find that $Kl(T)$ is symmetric monoidal. Of course, I may be making a mistake  here and I may edit this question with details of the verification later, but I think the most interesting part is about the diagonals.


First, I highlight that I'm using $\nabla$ to denote the morphism $\nabla: T(X) \otimes T(Y) \to T(X\otimes Y)$ from the condition of $T$ being monoidal.

We want a $\triangle_x^{Kl}$ in $Kl(T)$ such that $l_x^{Kl} \circ (e_x^{Kl} \otimes id_x^{Kl}) \circ \triangle_x^{Kl} = id_x^{Kl}$. If I'm understanding it correctly, those arrows are induced by the respective arrows in $C$, so $l_x^{Kl}$ in $Kl(C)$ corresponds to an arrow $\eta \circ l_x$ in $C$, $e_x^{Kl}$ to $\eta \circ e_x$ and $id_x^{Kl}$ to $\eta \circ id_x$.

Then the natural candidate for a diagonal in $Kl(T)$ is the composition $\eta \circ \triangle_x$. Using the composition in $Kl(T)$, $l_x^{Kl} \circ (e_x^{Kl} \otimes id_x^{Kl}) \circ \triangle_x^{Kl}$ is in $C$

$\require{AMScd}$
\begin{CD}
X @>\eta \circ \triangle_x>> T(X\otimes X) @>T(e_x\otimes id_x))>> T(I\otimes X) @>T(\eta\otimes \eta)>>T(T(I)\otimes T(X)) @>T(\nabla)>> TT(I\otimes X)@>\mu>>T(I\otimes X)@>T(l_x)>> T(X) @>T(\eta)>>TT(X) @>\mu>> T(X)
\end{CD}

Since $\mu \circ T(\eta) = id$ and $T(\eta \otimes \eta) \circ T(\nabla) = T(\eta)$ -- this holds because a coherence diagram of $T$ being monoidal says $\eta \otimes \eta \circ \nabla = \eta$ -- the above composition may be reduced to

$\require{AMScd}$
\begin{CD}
X @>\eta \circ \triangle_x>> T(X\otimes X) @>T(e_x\otimes id_x))>> T(I\otimes X)@>T(l_x)>> T(X) 
\end{CD}

Now, observe that the following diagram is commutative by the naturality of $\eta$


$\require{AMScd}$
\begin{CD}
X @>\triangle_x>> X\otimes X @>e_x\otimes id_x>>I\otimes X@>l_x>> X\\
@V\eta VV @V\eta VV  @V\eta VV  @V\eta VV \\T(X) @>T(\triangle_x)>> T(X\otimes X) @>T(e_x\otimes id_x))>> T(I\otimes X)@>T(l_x)>> T(X) 
\end{CD}

Since $\triangle_x$ is the diagonal in $C$, the first line in the above diagram is the identity and then we conclude that $\triangle_x^{Kl}$ is the diagonal in $Kl(C)$.  

Is this correct? Assuming it is, I have another question: the nLab states that ["For any cartesian monoidal category $C$ equipped with a monoidal monad $T$, the Kleisli category $Kl(T)$ is a Markov category".][2]

Since cartesian categories are symmetric and have diagonals, by my proof, $Kl(T)$ are symmetric and has diagonals. However, [every semicartesian category that is symmetric and had diagonal is cartesian][3]. Then, $Kl(C)$ is not an interesting Markov category because it is cartesian. 

What am I missing?  


  [1]: https://arxiv.org/pdf/1205.0101.pdf
  [2]: https://ncatlab.org/nlab/show/Markov+category
  [3]: https://mathoverflow.net/questions/348480/a-semicartesian-monoidal-category-with-diagonals-is-cartesian-proof