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1. • CommentRowNumber1.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2012
   • (edited May 17th 2012)
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   Author: Mirco Richter
   Format: MarkdownItexSuppose $V$ is a vector space and $C(V)$ its cofree coalgebra with a universal morphism $\epsilon$. Lets identify $C(V)$ for a moment with its underlaying vector space and take the cofree coalgebra $C^2(V) := C(C(V))$. Then $C^2(V) \simeq C(V)$ as cofree coalgebras. (To see that observe that there is a universal morphism $\sigma: C^2(V) \to C(V)$ and consequently a map $\delta:= \epsilon \circ \sigma$. To see that it is a universal morphism $C^2(V) \to V$ suppose $C$ is a coalgebra and $f:C \to V$ a linear map. Then there is a unique coalgebra map $g_1:C \to C(V)$ with $f=\epsilon\circ g_1$. In particular $g_1$ is a linear map and since $C^2(V)$ is the cofree coalgebra of $C(V)$ there is a unique coalgebra map $g_2:C \to C^2(V)$ with $g_1=\sigma\circ g_2$ and $f=\epsilon\circ \sigma\circ g_2 = \delta \circ g_2$.) -------------------------------------- In that case I would like to say, that the cofree coalgebra functor is 'idempotent'. But of course this can't be the right term since it is not an endofunctor. Moreover the proof is very basic and not at all specific to coalgebra. The only additional assumption (beside the cofreeness) is that the cofree object has an underlying object in the same category as the original object. Consequently this kind of 'idempotence' could be a property of many cofree functors. So my question: Is this a general property of cofree functors? And what is an adequate term for that kind of 'idempotence'?

   Suppose $V$ is a vector space and $C(V)$ its cofree coalgebra with a universal morphism $\epsilon$. Lets identify $C(V)$ for a moment with its underlaying vector space and take the cofree coalgebra $C^2(V) := C(C(V))$. Then

   $C^2(V) \simeq C(V)$ as cofree coalgebras.

   (To see that observe that there is a universal morphism $\sigma: C^2(V) \to C(V)$ and consequently a map $\delta:= \epsilon \circ \sigma$. To see that it is a universal morphism $C^2(V) \to V$ suppose $C$ is a coalgebra and $f:C \to V$ a linear map. Then there is a unique coalgebra map $g_1:C \to C(V)$ with $f=\epsilon\circ g_1$. In particular $g_1$ is a linear map and since $C^2(V)$ is the cofree coalgebra of $C(V)$ there is a unique coalgebra map $g_2:C \to C^2(V)$ with $g_1=\sigma\circ g_2$ and $f=\epsilon\circ \sigma\circ g_2 = \delta \circ g_2$.)

   ---

   In that case I would like to say, that the cofree coalgebra functor is ’idempotent’. But of course this can’t be the right term since it is not an endofunctor.

   Moreover the proof is very basic and not at all specific to coalgebra. The only additional assumption (beside the cofreeness) is that the cofree object has an underlying object in the same category as the original object. Consequently this kind of ’idempotence’ could be a property of many cofree functors.

   So my question: Is this a general property of cofree functors? And what is an adequate term for that kind of ’idempotence’?

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeMay 17th 2012
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   Author: Urs
   Format: MarkdownItex> this can't be the right term since it is not an endofunctor. If it's not an endofunctor, then you can't compose it with itself, as you did! So you are making a mistake somewhere. Do you see where? Hint: around the words "for a moment".

   this can’t be the right term since it is not an endofunctor.

   If it’s not an endofunctor, then you can’t compose it with itself, as you did!

   So you are making a mistake somewhere. Do you see where? Hint: around the words “for a moment”.

3. • CommentRowNumber3.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2012
   • (edited May 17th 2012)
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   Author: Mirco Richter
   Format: MarkdownItexI didn't it exactly because I forgot the coalgebra structure. But that is not a mistake. I just defined $C^2(V)$ as the cofree coalgebra of the vector space $C(V)$ and showed that this coalgebra is the same as the coalgebra $C(V)$.

   I didn’t it exactly because I forgot the coalgebra structure. But that is not a mistake. I just defined $C^2(V)$ as the cofree coalgebra of the vector space $C(V)$ and showed that this coalgebra is the same as the coalgebra $C(V)$.

4. • CommentRowNumber4.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2012
   • (edited May 17th 2012)
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   Author: Mirco Richter
   Format: MarkdownItexBut from your comment I guess, that the cofree coalgebra (lets call it $\tilde{C}^2(V)$ of the COFREE COALGEBRA $C(V)$ is not equivalent to $C(V)$ ... ? What I showed is just that the cofree coalgebra $C^2(V)$ of the VECTORSPACE $C(V)$ is equivalent to $C(V)$. Hence the underlying vector space structures of $\tilde{C}^2(V)$ and $C^2(V)$ are equivalent while (maybe) the coalgebra structures differs. Can we say that?

   But from your comment I guess, that the cofree coalgebra (lets call it $\tilde{C}^2(V)$ of the COFREE COALGEBRA $C(V)$ is not equivalent to $C(V)$ … ?

   What I showed is just that the cofree coalgebra $C^2(V)$ of the VECTORSPACE $C(V)$ is equivalent to $C(V)$. Hence the underlying vector space structures of $\tilde{C}^2(V)$ and $C^2(V)$ are equivalent while (maybe) the coalgebra structures differs. Can we say that?

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMay 17th 2012
   • (edited May 17th 2012)
   • PermaLink
   Author: Urs
   Format: MarkdownItexWhat I mean is that you are looking not just at the free functor from vector space to coalgebras, but secretly also at the [[forgetful functor]] back from coalgebras to their underlying vector spaces. While you did notice that you shoul "identitfy for a moment $C(V)$ with its underlying vector space" you need indeed to do that systematically, not just "for the moment". Precisely: You are really looking at a ([[free functor]] $\dashv $ [[forgetful functor]])-[[adjunction]] $$ Coalgebras \stackrel{\overset{Free}{\leftarrow}}{\underset{Forget}{\to}} VectorSpaces $$ and the operation that you are secretly applying is the composite of both, which is an [[endofunctor]] (namely the [[monad]] induced by the [[adjunction]]) $$ C : VectorSpaces \stackrel{Free}{\to} Coalgebras \stackrel{Forget}{\to} VectorSpaces \,. $$

   What I mean is that you are looking not just at the free functor from vector space to coalgebras, but secretly also at the forgetful functor back from coalgebras to their underlying vector spaces.

   While you did notice that you shoul “identitfy for a moment $C(V)$ with its underlying vector space” you need indeed to do that systematically, not just “for the moment”.

   Precisely:

   You are really looking at a (free functor $\dashv$ forgetful functor)-adjunction

   $$Coalgebras \stackrel{\overset{Free}{\leftarrow}}{\underset{Forget}{\to}} VectorSpaces$$

   and the operation that you are secretly applying is the composite of both, which is an endofunctor (namely the monad induced by the adjunction)

   $$C : VectorSpaces \stackrel{Free}{\to} Coalgebras \stackrel{Forget}{\to} VectorSpaces \,.$$
6. • CommentRowNumber6.
   • CommentAuthorMirco Richter
   • CommentTimeMay 17th 2012
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   Author: Mirco Richter
   Format: MarkdownItexYes ok I see. But a minor thing: it is forgetful functor $\dashv$ cofree functor.

   Yes ok I see. But a minor thing: it is forgetful functor $\dashv$ cofree functor.

7. • CommentRowNumber7.
   • CommentAuthorTobyBartels
   • CommentTimeMay 18th 2012
   • (edited May 18th 2012)
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   Author: TobyBartels
   Format: MarkdownItexI believe that you have an [[idempotent monad|idempotent]] [[comonad]].

   I believe that you have an idempotent comonad.

8. • CommentRowNumber8.
   • CommentAuthorUrs
   • CommentTimeMay 18th 2012
   • (edited May 18th 2012)
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   Author: Urs
   Format: MarkdownItexSure, my variance was the opposite.

   Sure, my variance was the opposite.