# Kleisli adjunction in a (weak) 2-category

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 $$K$$.

• How would a 'Kleisli object' be defined? – Berci Apr 30 '20 at 21:13
• @Berci I have added a link to the definition of a Kleisli object in my question. – Bob Apr 30 '20 at 21:44
• 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. – Kevin Arlin May 1 '20 at 6:24
• @KevinCarlson Following your hint, I have edited my question with my attempt at defining this adjunction. – Bob May 1 '20 at 19:49
• @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? – Bob May 2 '20 at 8:11

Hopefully I can expand on Kevin Arlin's comments in a helpful way.

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 $$\beta = \gamma\oppd f_t$$.

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.

• @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 $\alpha \operatorname{.}f_t : ff_t\to gf_t$. – jgon May 3 '20 at 20:57
• 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 $\lambda = \epsilon \operatorname{.}f_t$. – jgon May 3 '20 at 21:00
• @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. – jgon May 3 '20 at 22:10
• Thank you so much! In the morphism condition you wrote $\gamma : f \to h$. Don't you mean $\gamma : g \to h$? – Bob May 5 '20 at 19:40
• @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. – jgon May 14 '20 at 4:36