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Let $k$ be a commutative ring. There is a forgetful functor $$U : \mathsf{Coalg}_k \to \mathsf{Mod}_k$$ from $k$-coalgebras to $k$-modules. This has a right adjoint $R : \mathsf{Mod}_k \to \mathsf{Coalg}_k$, called the cofree coalgebra on a $k$-module. There are several ways to prove its existence, but let us assume that we have chosen an abstract approach (for example, by using some adjoint functor theorem). If necessary, let us assume that $k$ is a field. Can we derive from the universal property how the cofree coalgebra looks like explicitly?

So, given some $k$-module $M$ and a coalgebra $R(M)$ equipped with a natural bijection $\hom(C,R(M)) \cong \hom(U(C),M)$ for coalgebras $C$ , how can we make explicit what $R(M)$ is, i.e. what is the underlying $k$-module and its comultiplication? For example, by taking the coalgebra $C=(k \cdot x,\Delta(x)=x \otimes x,\varepsilon(x)=1)$, we see that $$\{a \in R(M) : \Delta(a)=a \otimes a, \, \varepsilon(a)=1\} \cong \hom(C,R(M)) \cong \hom(k \cdot x,M) \cong |M|.$$

Ender Wiggins
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Martin Brandenburg
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1 Answers1

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It does not follow exactly from the universal property, but on page 6 of this paper it is reported that in Richard E. Block, Pierre Leroux, "Generalized dual coalgebras of algebras, with applications to cofree coalgebras" it is shown that the cofree coalgebra of a vector space $V$ can be realized as the space of $k$-linear functions $f : k[t] \to T (V)$ (where $k[t]$ denotes the polynomial algebra and $T (V) = \bigoplus_{n=0}^{\infty} V^{\otimes n}$ the tensor algebra) that are of degree zero (i.e. $f(t^n) \in V^{\otimes n}$) and representative (i.e. for which there is a finite family $g_i, h_i: k[t] \to T (V)$ such that, for every $n, m \in \mathbb{N}$ we have $f(t^{nm}) = \sum_{i=1}^k g_i(t^n)h_i(t^m))$.

Ender Wiggins
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  • You claim (without giving any evidence) that my question has no answer, and then give a reference to an explicit construction of the cofree coalgebra, which is not what I ask about. – Martin Brandenburg Dec 30 '19 at 09:56
  • @MartinBrandenburg I beg your pardon if this upset you and if I didn't explain myself properly. I am not claiming anywhere that your question has no answer, instead I am providing a reference to an explicit construction that I recently found. This was aimed to help in two ways: firstly, you may now work with something concrete, instead of having an existence result obtained by applying an abstract approach. Secondly, if the authors found such an explicit description, maybe they started from the universal property, which means that the references may help in deriving the cofree from it. – Ender Wiggins Dec 30 '19 at 11:09
  • I am/was aware of explicit constructions. Again, this is not the content of my question. I only mentioned the abstract approach in my question to make clear that explicit constructions are exactly not what I would like to use. – Martin Brandenburg Dec 30 '19 at 14:25
  • Perhaps the following example explains better what I want: Assume you know that the coproduct $A \oplus B$ of abelian groups $A,B$ exists. Then you can actually derive the element structure of this abelian group, using the universal property (good exercise). It is not necessary to use an explicit construction. – Martin Brandenburg Dec 30 '19 at 14:27
  • @MartinBrandenburg I'm sorry but I don't understand what you mean by "element structure" and I would be glad to see how you can derive it from the universal property. I asked a question on the topic, to learn how to do it. What I may tell you is that, for me, the cofree coalgebra, as the other solutions to universal problems, is defined only up to isomorphism and hence, in a certain sense, any realization is derived by the universal property. It's just a matter of understanding how (or to write to the authors to ask for clarifications) – Ender Wiggins Dec 31 '19 at 10:01