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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 $f:A\to B^T$ 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 $f=\eta_Af^*$.


Questions:

  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?

Update:

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?

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A monad determines a reflective subcategory if and only if it is idempotent. That is well-known. –  Zhen Lin Mar 1 '13 at 13:14
    
Do you mean a full reflective subcategory? –  Berci Mar 1 '13 at 13:30
    
Reflective subcategories are always full in my definition. –  Zhen Lin Mar 1 '13 at 13:36
    
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 $TTA\cong TA$. –  Berci Mar 1 '13 at 13:40
    
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. –  askyle Jun 19 at 6:19
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2 Answers

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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.

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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)

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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? –  Berci Mar 2 '13 at 11:55
    
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. –  dorchard Mar 2 '13 at 16:41
    
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. –  dorchard Mar 2 '13 at 16:49
    
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. –  Berci Mar 2 '13 at 21:31
    
Whoops, yes I made a slight mistake here with the objects. –  dorchard Mar 7 '13 at 14:39
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