4
votes
0answers
138 views

Injectivity criterion for surjective coalgebra maps: does it hold in full generality?

Let $k$ be a commutative ring. Let $C$ be a filtered $k$-coalgebra. This means a $k$-coalgebra equipped with an increasing $k$-module filtration $C^0 \subseteq C^1 \subseteq C^2 \subseteq ...$ ...
1
vote
0answers
71 views

Coinduction and corestriction are quasi-inverse equivalences for comodules?

I'm reading http://arxiv.org/abs/math/0310337. There the following statement is given without proof: Let $k$ be a field. Let $C$ be a counitary coaugmented coalgebra, i.e. there is $\eta: C\to k$ ...
1
vote
1answer
243 views

(Co)Universal Property of Quotients/Subs

I'm not completely sure if this bunch of questions is the appropriate Level of MO. However at the same time I think that it is at least slightly above the level of stackex. ... The tensor algebra ...
1
vote
1answer
252 views

Reference for the fact that a coderivation of the (non reduced) tensor coalgebra is determined by its corestrictions

If $V$ is a vector space, let us consider the tensor coalgebra $TV=\bigoplus\limits_{k=0}^\infty V^{\otimes^k}$ with coproduct given by $$\Delta (x_1\otimes \dots \otimes x_n):= ...
1
vote
1answer
139 views

Linear functional kills all primitives of a connected filtered coalgebra => it lies in m^2?

Let $C$ be a connected filtered coalgebra over a field $k$. Maybe $k$ has characteristic $0$ (though I don't know where this can be of use). Let $1$ denote the unique element of $C_0$ mapping to $1\in ...
10
votes
1answer
527 views

Why not _co_free modules?

Let $R$ be a ring, and $R\text{-Mod}$ its category of all left modules. There is a "forgetful" functor $\operatorname{Forget}: R\text{-Mod} \to \text{AbGp}$, which is additive, continuous, and ...
17
votes
4answers
2k views

Is there an explicit construction of a free coalgebra?

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 ...