After answering this question here about Kleisli triples, I realized that this whole Kleisli triple construction:

$T:{\rm Ob}\mathcal C\to{\rm Ob}\mathcal C$, $\ \eta_A:A\to TA$ for all $A\in {\rm Ob}\mathcal C$ and $f\mapsto f^* $ for all $f:A\to TB$ such that

  1. $\eta_A^*=1_{TA}$
  2. $\eta_Af^*=f\ $ for all $\ f:A\to TB\ $ (writing composition from left to right)
  3. $(fg^*)^*=f^*g^*\ $ for all arrows $\ f:A\to TB,\ g:B\to TC$.

determines nothing else but a reflective subcategory $\ \tilde{\mathcal C}\ $ of $\ \mathcal C$ (consisting of exactly the arrows of the form $f^*$, among objects of the form $TA$):

The reflection of any $\ A\in{\rm Ob}\mathcal C\ $ to $\ \tilde{\mathcal C}$ is given by $\eta_A$, and then conditions 2. states exactly the reflection property (any $f$ from $A$ to $\tilde{\mathcal C}$, that is, any $f:A\to TB$ uniquely factors through $\eta_A$), and 1. and 3. ensure that $\tilde{\mathcal C}$ will be a subcategory.

Conversely, if a reflective subcategory $\tilde{\mathcal C}$ is given, we can fix a reflection arrow to $\tilde{\mathcal C}$ from each object $A\in{\rm Ob}\mathcal C$, and call them $\eta_A$, and its codomain $TA$. Then, as $TB\in{\rm Ob}\tilde{\mathcal C}$, for each $f:A\to TB$ we have a unique factorisation through $\eta_A$, and this determines $f^*$ so that $f=\eta_Af^*$.


  1. Is it a well know fact that monads thus are basically the same as reflective subcategories (at least up to natural isomorphism)?
  2. Is this argument deficient in any point?


Question 3. Starting out from a (not full) reflective subcategory $\mathcal B$ of $\mathcal C$, then constructing $\tilde{\mathcal C}$ by arbitrarily fixed reflection arrows $\eta_A$ as above, I can see that $\tilde{\mathcal C}\subseteq\mathcal B$ is a full reflective subcategory. Are they necessarily equivalent? If not, what more can we say about them?

  • 1
    $\begingroup$ A monad determines a reflective subcategory if and only if it is idempotent. That is well-known. $\endgroup$
    – Zhen Lin
    Mar 1, 2013 at 13:14
  • $\begingroup$ Do you mean a full reflective subcategory? $\endgroup$
    – Berci
    Mar 1, 2013 at 13:30
  • $\begingroup$ Reflective subcategories are always full in my definition. $\endgroup$
    – Zhen Lin
    Mar 1, 2013 at 13:36
  • $\begingroup$ So, that would be an easy consquence of the above, as $TA$ is already the reflection of $A$, and by fullness, $1_{TA}:TA\to TA$ is also reflection, so $TTA\cong TA$. $\endgroup$
    – Berci
    Mar 1, 2013 at 13:40
  • $\begingroup$ A question: given a Kleisli triple, how does one know that $\tilde{\mathcal{C}}$ is a subcategory? If $T$ is not injective one could have $f: A\to TB$ and $g: C\to TD$ with $TB=TC$, but no $h: A\to TD$ for which $h^* = f^*g*$. Morally it should be $h=\eta f^*g^*=fg^*$, but one can't use 3. to show that works because the types don't match. $\endgroup$
    – askyle
    Jun 19, 2014 at 6:19

2 Answers 2


I think I have found it.

So, the situation is indeed that each (not necessarily full) reflective subcategory determines a monad, and each monad determines a Kleisli triple that determines a reflective subcategory.

But, these procedures are not inverses to each other in both directions (they rather establish an adjoint situation only).

  1. If we start out from a monad $T:\mathcal C\to\mathcal C$, we construct $\tilde{\mathcal C}$, then consider its monad $\tilde T$, we get the same: $T\simeq \tilde T$ (naturally isomorphic).
  2. On the other hand, if we start out from a reflective subcategory $\mathcal B$ of $\mathcal C$, and fix reflections for each object $C\in{\rm OB}\mathcal C$ (that is, pick a left adjoint of the inclusion $\mathcal B\hookrightarrow\mathcal C$), and construct $\tilde{\mathcal C}$, we can well have that $\tilde{\mathcal C}$ is much smaller than (and not equivalent to) $\mathcal B$.

For an example, consider an arbitrary embedding $H:\mathcal{Grp}\hookrightarrow\mathcal{Set}$ (for example $H(G):=\{G\}\times UG$ where $UG$ is the underlying set of $G$), and let $\mathcal B$ be the image of $\mathcal{Grp}$ in $\mathcal{Set}$, consisting of the 'group homomorphisms' between these underlying sets.

Then, $\mathcal B$ is a reflective subcategory of $\mathcal{Set}$, and the reflection is the free group functor, so in this case $\widetilde{\mathcal{Set}}$ will consist only the ($H$-underlying sets of) free groups, which is strictly smaller than $\mathcal B$ itself.

However, in general, by the construction of $T$ for the reflective $\mathcal B\le\mathcal C$, we have $$\tilde{\mathcal C}\le \mathcal B\hookrightarrow \mathcal C^T$$ where $\mathcal C^T$ is the (Eilenberg-Moore) category of $T$-algebras, and the first containment is full and reflective.


Interesting. I think you are right. Let me attempt to flesh out your argument from another perspective based on the alternate definition of reflective subcategories: a subcategory $\tilde{\mathcal{C}} \subseteq \mathcal{C}$ is reflective in $\mathcal{C}$ when the inclusion functor $K : \tilde{\mathcal{C}} \rightarrow \mathcal{C}$ has a left adjoint $J : \mathcal{C} \rightarrow \tilde{\mathcal{C}}$ (MacLane).

Given a monad $T : \mathcal{C} \rightarrow \mathcal{C}$ (we'll work with Kleisli triple form here) there is a subcategory $\tilde{\mathcal{C}}$ where $Ob\tilde{\mathcal{C}} = Ob\mathcal{C}$ and for all $f \in \mathcal{C}(A, T B)$ then $f^\ast \in \tilde{\mathcal{C}}(A, B)$ ($\in \mathcal{C}(T A, T B)$).

The inclusion functor $K : \tilde{\mathcal{C}} \rightarrow \mathcal{C}$ is then defined $K A = T A$ on objects and $K f = f$ on morphisms.

Let's define a functor in the opposite direction: $J : \mathcal{C} \rightarrow \tilde{\mathcal{C}}$ where $J A = A$ on objects and $J f = (\eta \circ f)^\ast$ on morphisms (functoriality follows from monad laws).

This $J$ is left-adjoint to $K$ ($J \dashv K$) as there is a family of bijections $\phi : \tilde{\mathcal{C}}(J A, B) \cong \mathcal{C}(A, K B)$ (i.e. a subset of the bijections $\mathcal{C}(T A, T B) \cong \mathcal{C}(A, T B)$) defined $\phi f = f \circ \eta = f'^\ast \circ \eta$ (where $f'^\ast = f$ from the definition of $\tilde{\mathcal{C}}$) and $\phi^{-1} f = f^\ast$, where

$\phi^{-1} (\phi f) = (f'^\ast \circ \eta)^\ast = f'^\ast = f$

$\phi (\phi^{-1} f) = f^\ast \circ \eta = f$

(both following from the monad law: $g^\ast \circ \eta = g$).

Therefore $\tilde{\mathcal{C}}$ is a reflective subcategory of $\mathcal{C}$. $\Box$.

Conversley, given a reflective subcategory $\tilde{\mathcal{C}}$ with inclusion functor $K : \tilde{\mathcal{C}} \rightarrow \mathcal{C}$ then there is a left adjoint $J \dashv K$. Then we have a Kleisli triple ($KJ$, $\eta$, $K \phi^{-1}$) induced by the adjunction.

Cute result! Not sure how it is useful yet ;)

Btw, I think the idempotent monad result is that, the category of algebras for an idempotent monad is a reflective subcategory (little bit of information here: http://ncatlab.org/nlab/show/reflective+subcategory)

  • $\begingroup$ Thank you for your answer. This $\tilde{\mathcal C}$ (at least in the form you sketched), is rather the Kleisli category associated to $T$, and is not a subcategory of $\mathcal C$. But, I think that doesn't really matter.. However, there is still something which is not clear: if we start out from an arbitrary (but not full) reflective subcategory $\mathcal B\subseteq\mathcal C$, where not every $B\in{\rm Ob}\mathcal B$ is of the form $TC$ -- and, without fullness, $1_B$ is not guaranteed to be reflection -- Do we get back equivalent subcategory by this construction? $\endgroup$
    – Berci
    Mar 2, 2013 at 11:55
  • $\begingroup$ Thanks. Whilst the Kleisli category for $T$ is not a subcategory of $\mathcal{C}$ I do not think that $\tilde{\mathcal{C}}$ above is the Kleisli category, although it is isomorphic to it. Could you explain a bit more why you think $\tilde{\mathcal{C}}$ is the Kleisli category? Perhaps I am missing something. $\endgroup$
    – dorchard
    Mar 2, 2013 at 16:41
  • $\begingroup$ Regarding your second point, I was a little confused I'm afraid. In my answer I meant for the second (converse) part to be an arbitrary (non-full) reflective subcategory. If $1_B$ is the reflection (that is, the left adjoint to the inclusion) then doesn't this imply that $\tilde{\mathcal{B}}$ is equivalent to $\mathcal{C}$? Is that you're point. Sorry was a bit confused. $\endgroup$
    – dorchard
    Mar 2, 2013 at 16:49
  • $\begingroup$ I was also confused with your notation, like '$f^*\in\tilde{\mathcal C}(A,B)$' though the objects of $\tilde{\mathcal C}$ are of the form $TX$. For $1_B$ I meant the identity arrow $B\to B$, not a functor. But meanwhile, I guess I could also answer that part, too. All seems right. $\endgroup$
    – Berci
    Mar 2, 2013 at 21:31
  • $\begingroup$ Whoops, yes I made a slight mistake here with the objects. $\endgroup$
    – dorchard
    Mar 7, 2013 at 14:39

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