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In the 2-category of 1-categories, each monad $T$ on a category $\mathcal C$ determines a Kleisli category $\mathcal{C}_T$ and the so-called Kleisli adjunction between categories $\mathcal C$ and $\mathcal{C}_T$.

Let $\mathcal K$ be a (weak) 2-category, $a$ be a $0$-cell in $\mathcal K$, and $t$ be a monad on $a$. Assuming there exists a Kleisli object $a_t$, is there something like a Kleisli adjunction between $a$ and $a_t$? If yes, please describe it.

My attempt following the hint by @KevinCarlson: Assume that there is a Kleisli object $(f_t, \lambda)$. By applying $f_t$ to the identity map of the right $t$-module (that is $(a,t)$), you get a 1-cell from $a_t$ to $a$ that should be the right adjoint. Now, the unit should be $\lambda^{-1} \circ \eta$. But what is the counit? In Cat, the counit $\varepsilon_b$ would simply be $id_{t(b)}$ but I do not see how to generalize that to any 2-category $K$.

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  • $\begingroup$ How would a 'Kleisli object' be defined? $\endgroup$ – Berci Apr 30 '20 at 21:13
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    $\begingroup$ @Berci I have added a link to the definition of a Kleisli object in my question. $\endgroup$ – Bob Apr 30 '20 at 21:44
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    $\begingroup$ You get a map from $a_t\to a$ by applying the universal property to the identity map of the right $t$-module $(a,t)$. I assume this always gives the desired adjunction, building the unit and counit out of $\lambda$, although I’ve never checked this. $\endgroup$ – Kevin Arlin May 1 '20 at 6:24
  • $\begingroup$ @KevinCarlson Following your hint, I have edited my question with my attempt at defining this adjunction. $\endgroup$ – Bob May 1 '20 at 19:49
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    $\begingroup$ @KevinArlin Allow me to insist, so that I find the origin of the mistake you are pointing out: $F : C \to C_T$ maps $x$ to $x$ and $f : x \to y$ to $\eta_Y \circ f : x \to T y$. $U : C_T \to C$ maps $x$ to $T x$ and $f : x \to T y$ to $\mu_Y \circ T f : Tx \to Ty$. Therefore $\varepsilon_x$ goes from $FUx$ to $Tx$, that is from $Tx$ to $Tx$. Where am I wrong? $\endgroup$ – Bob May 2 '20 at 8:11
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Hopefully I can expand on Kevin Arlin's comments in a helpful way.

The answer is yes, there is a Kleisli adjunction.

Preamble

I'll copy the nLab definitions to stay self-contained.

Let $K$ be a 2-category, $t:a\to a$ a monad, $(a_t,f_t,\lambda)$ a Kleisli object for $t$, meaning a representing object for the functor $K\to \newcommand\Cat{\mathbf{Cat}}\Cat\newcommand\oppd{\operatorname{.}}$ that sends an object $x$ to the right $t$-modules on $x$, $\newcommand\RMod{\operatorname{RMod}}\RMod(x,t)$. So $a_t$ is a 0-cell, $f_t:a\to a_t$ a 1-cell, and $\lambda:f_tt\to f_t$ a 2-cell, such that for any right module $(r,\alpha)$, with $r:a\to x$, $\alpha : rt\to r$, there is a unique morphism $a_t\to x$ whose composite with $f_t$ (resp. $\lambda)$ is $r$ (resp. $\alpha$).

Edit: An elementary reformulation of the definition of a Kleisli object:

A Kleisli object for a monad $(a,t:a\to a,\mu:t^2\to t,\eta : 1_a\to t)$ consists of the data of a 0-cell $a_t$, and a right $t$-module $(f_t : a\to a_t,\lambda : f_tt\to f_t)$ on $a_t$ such that the following universality conditions are satisfied.

Object condition: For any right $t$-module on $x$, $(r:a\to x, \alpha : rt\to r)$, there is a unique morphism $g : a_t\to x$ such that $(r,\alpha) = (gf_t, g\oppd \lambda)$.

Morphism condition: For two right $t$-modules on $x$, which we know are of the form $(gf_t,g\oppd\lambda)$ and $(hf_t,h\oppd\lambda)$ by the object condition, for $g,h:a_t\to x$ and for every morphism of right $t$-modules $\beta: gf_t\to hf_t$, there is a unique 2-cell $\gamma : g\to h$ such that $\beta = \gamma\oppd f_t$.

The adjoint 1-cells:

We already have $f_t:a\to a_t$, so we need $g_t:a_t\to a$, which should correspond to a right $t$-module structure on $a$. Luckily, we already have a canonical one, $(t,\mu)$, where $\mu:t^2\to t$ is the multiplication of the monad. Thus we get a map $g_t$ from the universal property, such that $g_tf_t=t$ and $g_t\oppd\lambda = \mu$.

The unit:

Then the unit of the monad, $\eta:1_a\to t=g_tf_t$ is the unit of the adjunction.

Constructing the counit:

To construct the counit, $\epsilon : f_tg_t\to 1_{a_t}$, we need to understand $f_tg_t : a_t\to a_t$. However, since $a_t$ represents right modules, this morphism classifies the right module on $a_t$, $(f_tg_tf_t,f_tg_t\oppd\lambda)$, but by definition of $g_t$, this is equal to $(f_tt,f_t\oppd\mu)$.

Similarly, $1_{a_t}$ corresponds to the module $(f_t,\lambda)$.

Now you can check that $\lambda: f_tt\to f_t$ is a morphism of right $t$-modules between these two, since $$ \require{AMScd} \begin{CD} f_ttt @>f_t\oppd\mu>> f_tt \\ @V\lambda\oppd t VV @VV\lambda V\\ f_tt @>\lambda>> f_t \\ \end{CD} $$ commutes, because this diagram is one of the diagrams that are required for $\lambda$ to be a multiplication making $f_t$ a $t$-module in the first place.

Thus $\lambda$ induces a morphism $\epsilon : f_tg_t\to 1_{a_t}$ satisfying $\epsilon\oppd f_t = \lambda$.

The triangle identities:

For the triangle identities, we now have $$(\epsilon\oppd f_t)(f_t\oppd \eta) = \lambda(f_t.\eta)=1_{f_t}$$ by the unit axiom of $\lambda$. For the other, we can understand $$(g_t\oppd \epsilon)(\eta \oppd g_t) : g_t\to g_t $$ by composing with $f_t$ to get the corresponding endomorphism of the right $t$-module $(t,\mu)$. $$((g_t\oppd \epsilon)(\eta\oppd g_t))\oppd f_t = (g_t\oppd \epsilon \oppd f_t)(\eta\oppd g_t\oppd f_t) = (g_t\oppd \lambda)(\eta\oppd t) = \mu(\eta\oppd t) = 1_t, $$ by the unit axiom of $\mu$. Since $1_t = 1_{g_t}\oppd f_t$, we conclude $$(g_t\oppd \epsilon)(\eta\oppd g_t) = 1_{g_t},$$ as desired.

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    $\begingroup$ @Bob Finally, for the definition of $\epsilon$, part of what it means for $a_t$ to represent the right $t$-module functor is that morphisms of right $t$-modules (on some object $x$) need to correspond to 2-cells between the corresponding morphisms $a_t\to x$. Given $f,g : a_t\to x$, and $\alpha : f\to g$ a 2-cell, the corresponding right modules are $(x,ff_t,f\operatorname{.}\lambda)$ and $(x,gf_t,g\operatorname{.}\lambda)$, and the morphism of $t$-modules is given by $\alpha \operatorname{.}f_t : ff_t\to gf_t$. $\endgroup$ – jgon May 3 '20 at 20:57
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    $\begingroup$ Since $a_t$ is a representing object, we also have a converse. Any morphism of $t$-modules $(x,ff_t,f.\lambda)\to (x,gf_t,g.\lambda)$ is of the form $\alpha.f_t$ for a unique $2$-cell $\alpha: f\to g$. Therefore $\epsilon$ is defined to be the unique morphism satisfying $\lambda = \epsilon \operatorname{.}f_t$. $\endgroup$ – jgon May 3 '20 at 21:00
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    $\begingroup$ @Bob The elementary reformulation is incomplete on nLab. I'll edit in a complete elementary reformulation into my top section. nLab is only talking about what happens with 1-cells, and not what happens to 2-cells. $\endgroup$ – jgon May 3 '20 at 22:10
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    $\begingroup$ Thank you so much! In the morphism condition you wrote $\gamma : f \to h$. Don't you mean $\gamma : g \to h$? $\endgroup$ – Bob May 5 '20 at 19:40
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    $\begingroup$ @Bob, yes I definitely wanted to avoid being careful about about whether I had a strict 2-category or not, which in principle should be fine, since we can strictify a non strict 2-category. The non strict version presents a fair bit of extra complexity, not least of all because nLab unhelpfully chooses not to define such a notion. I have a guess as to what the correct notion is, but I've been a bit busy recently, so I don't have time to check right now. My best guess right now is just that we replace equality with isomorphism in the object condition part of the definition. $\endgroup$ – jgon May 14 '20 at 4:36

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