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Is “free” a slip for cofree? Generally speaking, “free” refers to a left adjoint to a forgetful functor, but the forgetful functor from coalgebras to vector spaces doesn’t have a left adjoint. It does have a right adjoint, called the cofree coalgebra on a vector space.
There is not much good discussion on this (yet) in the nLab; there’s a little bit here, and some over at linear logic. If you apply the search function within the nLab to “cofree”, you may find pages closer to your interests (e.g., differential graded coalgebra). I mean to link back to (maybe even amplify on) the MathOverflow discussion I think you’re referring to; Theo Johnson-Freyd’s answer is IMO very nice.
By the way, I didn’t mean to imply that “free” was your slip; indeed, I remember it was the OP of the MO discussion who introduced the slip. And I’m afraid I really do consider it a mistake, or at the very least highly misleading. I do not however insist on coassociative or cocommutative unless it’s important to do so (e.g., in a Hopf algebra context). There is, by the way, a notion of cotensor = power in category theory, and also a notion of cotensor product. :-)
Started something on cofree coalgebra.
I have added various hyperlinks.
Also, I have added cross-links between this entry and the formal duals free monoid and tensor algebra.
It is usual to say coassociative coalgebra, as usually one deals with bialgebra and similar contexts where both algebra and coalgebra structures are present so it is good to emphasise which “associativity” one means. But of course, it is OK in pure coalgebra context to say associativity for dual associativity. Of course, one can not say cotensor coalgebra for tensor coalgebra; as it is based on the tensor and not on the cotensor product.
I have been adding a bit more to cofree coalgebra. In the course of doing so, I’ve pondered a bit more on the construction given by Theo in the MathOverflow discussion alluded to in #4, and I don’t quite see how his description (as a certain pullback in $Vect$, if I’m reading his description correctly) gives a coalgebra. In particular, I don’t understand how the comultiplication for his description is supposed to work.
If the inclusion of the pullback $V \times_{V^{\ast \ast}} T(V^\ast)^\circ \to T(V^\ast)^\circ$ gives a subcoalgebra inclusion, then I believe all would be well. But at the moment, I have my suspicions that the pullback is “too big” for this to actually work. In any case, I’m hoping Theo can address this here.
I also realize that what I’ve written up at cofree coalgebra is unlikely to satisfy Jim (see #1); similar discussions for cofree cocommutative coalgebras took place at the Café some years ago, but I’m pretty sure they were not carried out to Jim’s full satisfaction. I hope a better job can be done this time around, but there is rather a lot to say.
Yes, that accords exactly with what I was thinking. Thanks for getting back to me on this, Theo.
Interesting. The present version of the story as recorded at cofree coalgebra is just to say that $Cofree(V) = T(V^\ast)^\circ$ when $V$ is finite-dimensional, and in general $Cofree(V)$ is the filtered colimit (or you could simply say, “union”) over $Cofree(V')$ where $V'$ ranges over finite-dimensional subspaces of $V$ and inclusions between them (NB: this colimit can be computed in $Vect$). This translates easily into your revised description at MathOverflow.
Incidentally, looking at your direct sum description of the cofree (cocommutative) coalgebra on a 1-dimensional space $k$ (which we may identify with its dual), I would be interested to see exactly how it embeds as a subspace of $T(k)^\ast \cong \prod_n k \cdot x^n$, the vector space of formal power series in one variable. My own description of $T(k)^\circ \hookrightarrow T(k)^\ast$ is that it is (the underlying space of) the localization of $k[x]$ at the prime ideal $(x)$, embedded by writing each rational function in the localization as a formal power series. I discussed this calculation quite a while ago here. Our intuitions about finitely supported distributions have something in common, but I’m not quite seeing why the coalgebra is splitting so neatly as a direct sum of copies of $T(k)$ as you are claiming (even assuming the field $k$ is algebraically closed).
My most recent edit at cofree coalgebra answers the question I raised in comment #13, in what could be an embarrassment of detail. Theo’s description of the cofree coalgebra on one cogenerator is correct, but the nitty-gritty details are perhaps not super-trivial. (Still more to be done at that article.)
Jim, it is very good to hear from you. I hope you are well.
The free commutative $\mathbb{Z}$-algebra on one element is the polynomial algebra $\mathbb{Z}[x]$. The free commutative $\mathbb{Z}$-algebra on $n$ elements is the polynomial algebra on $n$ variables, $\mathbb{Z}[x_1, \ldots, x_n]$.
I feel a little shaky about cofree coalgebras when the base ring is not a field, but let me try to get back to you later. There’s some related discussion in this thread, including discussion about the usage of “free” versus “cofree”. In the past, you’ve never seemed happy with my attempts to describe cofree coalgebras. :-) :-(
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