symmetric monoidal (∞,1)-category of spectra
Given a monoidal category $M$, let $Coalg(M)$ denote the category of coalgebras (i.e., comonoids) with respect to the tensor product of $M$. Let $U \colon Coalg(M) \to M$ be the forgetful functor. The cofree coalgebra refers to a functor right adjoint to $U$, if it exists.
For example, a cofree coalgebra over a vector space $V$ consists of a coalgebra $C(V)$ and a linear map $\varepsilon \colon C(V) \to V$ which is universal in the sense that given a coalgebra $C$ and linear map $f \colon C \to V$, there exists a unique coalgebra map $g \colon C \to C(V)$ such that $f = \varepsilon \circ g$.
The forgetful functor $U \colon Coalg(M) \to M$ preserves and reflects (creates) any colimits which happen to exist, analogous to the fact that the forgetful functor $Alg(M) \to M$ preserves and reflects limits. Therefore, under hypotheses of an adjoint functor theorem (which almost always obtain in cases of practical interest), $U$ will indeed have a right adjoint.
The special adjoint functor theorem (SAFT) may be applied to show that for any commutative ring $K$, the forgetful functor
has a right adjoint. An argument is given by Michael Barr in
(This is somewhat at odds with an assertion made by Michiel Hazewinkel in Witt Vectors, Part I, section 12.11 (page 58): “Whether the cofree coalgebra over an Abelian group always exists is unknown.” But it seems to me (Todd Trimble) that there is nothing wrong with Barr’s argument.)
In what follows, we give an explicit construction of the cofree coalgebra cogenerated by a vector space.
As a first step, recall the following result.
Every coalgebra is the filtered colimit over the diagram of its finite-dimensional subcoalgebras and inclusions between them.
For a proof of this result, see for example Michaelis. We repeat that colimits of coalgebras are created by taking colimits of their underlying vector spaces.
Let for a vector space $V$, let $T(V)$ be the tensor algebra (or free algebra) generated by $V$. We may consider $T(V^\ast)$ as the algebra of (non-commutative) polynomial functions on $V$. The dual of $T(V^\ast)$ is not a coalgebra, but in accordance with the fundamental theorem above, we may consider instead the colimit over a system of finite-dimensional coalgebras obtained by dualizing the system of finite-dimensional algebra quotients of $T(V^\ast)$ and surjections between them (compare profinite completion), using the fact that the dual of a finite-dimensional algebra is a finite-dimensional coalgebra. Let $T(V^\ast)^\circ$ be this colimit coalgebra, embedded in $T(V^\ast)^\ast$. The dual of the inclusion $V^\ast \hookrightarrow T(V^\ast)$ gives a map $T(V^\ast)^\ast \to V^{\ast \ast}$, and this restricts to a map $T(V^\ast)^\circ \to V^{\ast \ast}$.
For finite-dimensional $V$, there is an isomorphism $V^{\ast \ast} \cong V$, and we have the following result.
If $V$ is finite-dimensional, then the composite $T(V^\ast)^\circ \to V^{\ast \ast} \cong V$ exhibits $T(V^\ast)^\circ$ as the cofree coalgebra over $V$.
Suppose given a coalgebra $C$ and a linear map $f \colon C \to V$. We want to show this has a unique lift to a coalgebra map $\widehat{f} \colon C \to T(V^\ast)^\circ$. For $x \in C$, there is a minimal finite-dimensional subcoalgebra $C_x$ containing $x$, and if $f_x \colon C_x \to T(V^\ast)^\circ$ is the restriction of $f$, then the value $\widehat{f}(x)$ must match the value $\widehat{f_x}(x)$. In other words, in view of the fundamental theorem of coalgebras, there is no loss of generality in taking $C$ to be finite-dimensional.
In this case, the map $f^\ast \colon V^\ast \to C^\ast$ extends uniquely to an algebra map $T(V^\ast) \to C^\ast$, since the tensor algebra is the free algebra on $V$. This algebra map factors as
where $Q$ is a finite-dimensional algebra. The dual $Q^\ast$ carries a coalgebra structure that embeds in $T(V^\ast)^\circ$, and by dualizing we obtain coalgebra maps
that provide the desired lift. The uniqueness of the lift is evident from the uniqueness of the extension $T(V^\ast) \to C$ (or equivalently, of $T(V^\ast)_{prof} \to C$, passing to the profinite completion) on the dual algebra side.
Now let $V$ be an infinite-dimensional vector space. We may view $V$ as the union or filtered colimit over the system $\{V_\alpha\}$ of its finite-dimensional subspaces. Applying the cofree coalgebra construction to these finite-dimensional subspaces, we get a system of coalgebras.
The coalgebra colimit
is the cofree coalgebra over $V$.
Indeed, suppose given a coalgebra $C$ and a linear function $f \colon C \to V$. To construct a coalgebra map $C \to K$ which lifts $f$, we may argue just as we did in the proof of the lemma, and suppose without loss of generality that $C$ is finite-dimensional. Then of course the map $f \colon C \to V$ factors through a map $f_\alpha \colon C \to V_\alpha$ where $V_\alpha$ is a finite-dimensional subspace of $V$. This lifts uniquely to a coalgebra map $\widehat{f_\alpha} \colon C(V_\alpha)$, and the desired lift $\widehat{f}$ is the composite
Any such lift is obtained in just this way.
It is easy to see that this description is equivalent to that described in a MathOverflow discussion here. See also the nForum discussion here.
Specializing still further, we investigate the detailed structure of the cofree coalgebra over a 1-dimensional vector space (over a field) $k$.
Here, the cofree coalgebra $T(k)^\circ$ is the filtered colimit of finite-dimensional coalgebras of the form
where $k[x]/I$ is a (finite-dimensional) quotient of the polynomial algebra $k[x]$ modulo a non-zero ideal $I$. The adjoint of the quotient map $k[x] \to k[x]/I$ is an embedding $(k[x]/I)^\ast \hookrightarrow k[x]^\ast \cong \prod_n k \cdot x^n$ into the space of formal power series in $x$. As a set, $T(k)^\circ$ is simply the union of the subspaces $(k[x]/I)^\ast \hookrightarrow \prod_n k \cdot x^n$. In other words, every $\xi \in T(k)^\circ$ belongs to some $(k[x]/I)^\ast$, and it suffices to consider first the structure of a typical such $\xi$ as a formal power series, and then how the comultiplication behaves on it.
Now $I$ is a principal ideal $(g(x))$ for some polynomial $g(x) = b_0 + \ldots + b_n x^n$, where the top coefficient is of course assumed non-zero. We are trying to understand when a formal power series $\xi = \sum_{n \geq 0} a_n x^n \in \prod_n k \cdot x^n$ vanishes on $I$, in the sense that
for all $p(x) \in (g(x))$, where the pairing $\langle , \rangle \colon k[x] \times \prod_n k \cdot x^n \to k$ is the canonical one:
Taking the case $p(x) = x^j g(x)$, we obtain recurrence relations
one for every $j$. We may pick $a_0, a_1, \ldots, a_{n-1}$ at random, but from there on the remainder of the sequence $a_n, a_{n+1}, \ldots$ is determined by the recurrence relations.
Now let $g^{rev}(x)$ be the polynomial obtained by reversing the order of the coefficients $b_0, \ldots, b_n$ of $g$,
and consider the problem of multiplicatively inverting $g^{rev}(x)$ (which we may do since the constant coefficient of $g^{rev}$ is non-zero). If $\sum_j a_j x^j$ is the reciprocal, we have
which yields a second set of recurrence relations by examining coefficients of powers of $x$. We get the first set of recurrence relations by restricting attention to the coefficients of $x^n$, $x^{n+1}$, etc.
Thus the power series of $g^{rev}(x)^{-1}$ yields a $\xi$ which vanishes on $I = (g(x))$, and similarly the power series of
also vanish on $I$. They in fact form a basis for the subspace $(k[x]/(g(x)))^\ast \hookrightarrow \prod_n k \cdot x^n$.
We conclude that an arbitrary element of $T(k)^\circ$ is of the form $\frac{p(x)}{h(x)}$ for some $h$ with non-zero constant coefficient:
As subspaces of the space of formal power series $\prod_n k \cdot x^n$, there is an identification
where the right side is the localization of $k[x]$ at the prime ideal $(x)$, whose elements are considered as formal power series.
It remains to identify the coalgebra structure on $k[x]_{(x)}$. This extends the standard coalgebra structure on the tensor algebra $k[x]$, where the comultiplication is given by deconcatenation:
We may as well focus on a typical subcoalgebra $(k[x]/(g(x)))^\ast$, with basis elements $\frac{x^i}{g^{rev}(x)}$. By duality, the definition of $\delta(\frac{x^i}{g^{rev}(x)})$ can be extracted from the multiplication table for residue classes $x^j \pmod g(x)$.
The calculations work out cleanest if we assume if $k$ is algebraically closed. For in that case, we can take advantage of partial fraction decompositions of the form
so that it suffices to give $\delta(\frac1{(1 - r x)^n})$, although the algebra works out better with a slightly different basis.
With this in mind, put $e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}}$ for $n \geq 0$. The power series expansion is
and we formally calculate
which exhibits comultiplication as deconcatenation, familiar from the usual coalgebra structure on $k[x]$. In particular, we have $\delta(e_0(r)) = e_0(r) \otimes e_0(r)$: the elements $e_0(r)$ are grouplike.
A more respectable-looking calculation (but really with the exact same content!) shows that the comultiplication thus defined on $k[x]_{(x)}$ is indeed adjoint to multiplication on $k[x]$:
An analogous calculation shows that the counit $\varepsilon \colon k[x]_{(x)} \to k$ is defined by $\varepsilon(e_n(r)) = \delta_{0 n}$ (Kronecker delta).
We summarize the calculations above as follows.
Let $k$ be an algebraically closed field. For each nonzero scalar $r \in k$, let $L(r)$ denote the $k$-linear span of the elements $e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}}$ in $k[x]_{(x)}$, with coalgebra structure given in each case by
Let $L(0) = k[x]$ as a subspace of $k[x]_{(x)}$. The cofree coalgebra over a 1-dimensional space $k$, identified with $k[x]_{(x)}$, decomposes as a direct sum of coalgebras
according to the partial fraction decomposition of rational functions. Each of the coalgebras $L(r)$ is isomorphic to $k[x]$ with its standard coalgebra structure.
See also
but note that this contains an error which is corrected in