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A monad $T$ on $\cal C$ is a lax functor $\mathbb T : \mathbf{1}\to\bf Cat$. The lax colimit and limit of $\mathbb T$ are the Klesili and EM categories ${\cal C}^{\mathbb T}$, ${\cal C}_{\mathbb T}$ of the monad.

Hence, it seems reasonable for the category of lax co/cones for $\mathbb T$ to be equivalent to the category whose objects are adjunctions $F\dashv G$ splitting the given monad as a composition $GF$.

Writing down the definition of a lax cone for $\mathbb T$, one obtains a functor $S\colon {\cal X}\to \cal C$ with a natural transformation $\sigma \colon TS\to S$ satisfying certain commutativities: $$ \begin{array}{lr} \begin{array}{ccc} TTS &\to & TS \\ \downarrow && \downarrow \\ TS &\to& S \end{array} & \begin{array}{ccc} S &\to &TS\\ &\searrow&\downarrow\\ && S \end{array}\\ &&\\ \sigma\circ \mu S = \sigma\circ T\sigma, & \sigma \circ \eta S = 1_S \end{array} $$ This means that morally $\sigma = S\epsilon$, for $\epsilon\colon LS\to 1$ the counit of the wannabe adjunction (the equality $\sigma \circ \eta S = 1_S$ is hence half of the zigzag identities).

It seems now unlikely that the functor $S\colon \cal X\to C$, the right adjoint in the wannabe adjunction $L\dashv S$, becomes automatically continuous. So,

is the conjecture above false? What should it be replaced by?

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I believe that your claim is false.

Consider $\mathcal X=1$ then a pair $\langle S,\sigma \rangle$ as you describe (a cone for $\mathbb T$) should be nothing but:

  • an object $S(*)$, where $*$ is the only object in $1$;
  • a $T$-algebra structure over $S(*)$, namely $\sigma_* \colon TS(*) \to S(*)$.

The left adjoint for $S$ should be an $L \colon \mathcal C \to 1$ and there are not many functors of this kind, just the constant one.

Similarly $\epsilon \colon LS \to 1_{1}$, the counit, is given by the only morphism $\epsilon_* \colon LS(*)=* \to *$ which must be $1_{*}$, the only morphism in $1$.

If your claim was true, that is $S\epsilon=\sigma$, this would imply that $$\sigma_*=S(\epsilon_*)=S(1_*)=1_{S(*)}\ .$$

This is basically saying that every $T$-algebra should have the identity as underlying structure map. This is not clearly the case because:

  • for start there is no reason why $TS(*)=S(*)$ ($TS(*)$ is the source of $\sigma$, while $S(*)$ is the source of $1_{S(*)}$);
  • there are lots of monads whose algebras have structure maps that are not the identity.
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