TITLE: Is there an explicit construction of a free coalgebra?
QUESTION [36 upvotes]: I am interested in the differences between algebras and coalgebras. Naively, it does not seem as though there is much difference: after all, all you have done is to reverse the arrows in the definitions. There are some simple differences: 
The dual of a coalgebra is naturally an algebra but the dual of an algebra need not be naturally a coalgebra.  
There is the Artin-Wedderburn classification of semisimple algebras. I am not aware of a classification even of simple, semisimple coalgebras.  
More surprising is: a finitely generated comodule is finite dimensional.
This question is about a more striking difference. The free algebra on a vector space $V$ is $T(V)$, the tensor algebra on $V$. I have been told that there is no explicit construction of the free coalgebra on a vector space. However these discussions took place following the consumption of alcohol. What is known about free coalgebras?

REPLY [2 votes]: My two cents on the topic: dealing with co-structures, conilpotency (sometimes called connectedness or even cocompleteness) is a necessary condition to talk about cofreeness. A known counterexample is given by the coalgebra $(\mathbb k, \Delta),$ with $\Delta(1)=1\otimes 1$, denoting by $\mathbb k$ a field. The coalgebra is clearly not cocomplete and does not satisfy the lifting property of coalgebra morphisms (considering reduced tensor coalgebras, which are cocomplete).
I would like to share two references:

The book Algebraic Operads  by Loday and Vallette describes the topic in much detail (see, in particular, pag.8-9 in Chapter I).
The first chapter of Hasegawa's PhD thesis contains the definition of cocomplete coalgebras and discusses lifting properties.