cofree coalgebra



Given a monoidal category MM, let Coalg(M)Coalg(M) denote the category of coalgebras (i.e., comonoids) with respect to the tensor product of MM. Let U:Coalg(M)MU \colon Coalg(M) \to M be the forgetful functor. The cofree coalgebra refers to a functor right adjoint to UU, if it exists.


For example, a cofree coalgebra over a vector space VV consists of a coalgebra C(V)C(V) and a linear map ε:C(V)V\varepsilon \colon C(V) \to V which is universal in the sense that given a coalgebra CC and linear map f:CVf \colon C \to V, there exists a unique coalgebra map g:CC(V)g \colon C \to C(V) such that f=εgf = \varepsilon \circ g.


The forgetful functor U:Coalg(M)MU \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)MAlg(M) \to M preserves and reflects limits. Therefore, under hypotheses of an adjoint functor theorem (which almost always obtain in cases of practical interest), UU will indeed have a right adjoint.

Modules over a commutative ring

The special adjoint functor theorem (SAFT) may be applied to show that for any commutative ring KK, the forgetful functor

K-CoalgK-ModK\text{-Coalg} \to K\text{-Mod}

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.)

Cofree coalgebra over a vector space

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.

Fundamental theorem of coalgebras

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 VV, let T(V)T(V) be the tensor algebra (or free algebra) generated by VV. We may consider T(V *)T(V^\ast) as the algebra of (non-commutative) polynomial functions on VV. The dual of T(V *)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 *)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 *) T(V^\ast)^\circ be this colimit coalgebra, embedded in T(V *) *T(V^\ast)^\ast. The dual of the inclusion V *T(V *)V^\ast \hookrightarrow T(V^\ast) gives a map T(V *) *V **T(V^\ast)^\ast \to V^{\ast \ast}, and this restricts to a map T(V *) V **T(V^\ast)^\circ \to V^{\ast \ast}.

For finite-dimensional VV, there is an isomorphism V **VV^{\ast \ast} \cong V, and we have the following result.


If VV is finite-dimensional, then the composite T(V *) V **VT(V^\ast)^\circ \to V^{\ast \ast} \cong V exhibits T(V *) T(V^\ast)^\circ as the cofree coalgebra over VV.


Suppose given a coalgebra CC and a linear map f:CVf \colon C \to V. We want to show this has a unique lift to a coalgebra map f^:CT(V *) \widehat{f} \colon C \to T(V^\ast)^\circ. For xCx \in C, there is a minimal finite-dimensional subcoalgebra C xC_x containing xx, and if f x:C xT(V *) f_x \colon C_x \to T(V^\ast)^\circ is the restriction of ff, then the value f^(x)\widehat{f}(x) must match the value f x^(x)\widehat{f_x}(x). In other words, in view of the fundamental theorem of coalgebras, there is no loss of generality in taking CC to be finite-dimensional.

In this case, the map f *:V *C *f^\ast \colon V^\ast \to C^\ast extends uniquely to an algebra map T(V *)C *T(V^\ast) \to C^\ast, since the tensor algebra is the free algebra on VV. This algebra map factors as

T(V *)ontoQintoC *T(V^\ast) \stackrel{onto}{\to} Q \stackrel{into}{\to} C^\ast

where QQ is a finite-dimensional algebra. The dual Q *Q^\ast carries a coalgebra structure that embeds in T(V *) T(V^\ast)^\circ, and by dualizing we obtain coalgebra maps

CQ *T(V *) C \to Q^\ast \to T(V^\ast)^\circ

that provide the desired lift. The uniqueness of the lift is evident from the uniqueness of the extension T(V *)CT(V^\ast) \to C (or equivalently, of T(V *) profCT(V^\ast)_{prof} \to C, passing to the profinite completion) on the dual algebra side.

Now let VV be an infinite-dimensional vector space. We may view VV as the union or filtered colimit over the system {V α}\{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

K=colim V αVC(V α)K = colim_{V_\alpha \subset V} C(V_\alpha)

is the cofree coalgebra over VV.

  • Remark: Happily, this colimit is computed as in VectVect, and more happily still – since the underlying functor VectSetVect \to Set also preserves and creates filtered colimits – this colimit is just a set-theoretic union of the C(V α)C(V_\alpha) along the canonical inclusions between them.

Indeed, suppose given a coalgebra CC and a linear function f:CVf \colon C \to V. To construct a coalgebra map CKC \to K which lifts ff, we may argue just as we did in the proof of the lemma, and suppose without loss of generality that CC is finite-dimensional. Then of course the map f:CVf \colon C \to V factors through a map f α:CV αf_\alpha \colon C \to V_\alpha where V αV_\alpha is a finite-dimensional subspace of VV. This lifts uniquely to a coalgebra map f α^:C(V α)\widehat{f_\alpha} \colon C(V_\alpha), and the desired lift f^\widehat{f} is the composite

Cf α^C(V α)KC \stackrel{\widehat{f_\alpha}}{\to} C(V_\alpha) \hookrightarrow K

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.

The cofree coalgebra over a 1-dimensional space

Specializing still further, we investigate the detailed structure of the cofree coalgebra over a 1-dimensional vector space (over a field) kk.

Here, the cofree coalgebra T(k) T(k)^\circ is the filtered colimit of finite-dimensional coalgebras of the form

(k[x]/I) *(k[x]/I)^\ast

where k[x]/Ik[x]/I is a (finite-dimensional) quotient of the polynomial algebra k[x]k[x] modulo a non-zero ideal II. The adjoint of the quotient map k[x]k[x]/Ik[x] \to k[x]/I is an embedding (k[x]/I) *k[x] * nkx n(k[x]/I)^\ast \hookrightarrow k[x]^\ast \cong \prod_n k \cdot x^n into the space of formal power series in xx. As a set, T(k) T(k)^\circ is simply the union of the subspaces (k[x]/I) * nkx n(k[x]/I)^\ast \hookrightarrow \prod_n k \cdot x^n. In other words, every ξT(k) \xi \in T(k)^\circ belongs to some (k[x]/I) *(k[x]/I)^\ast, and it suffices to consider first the structure of a typical such ξ\xi as a formal power series, and then how the comultiplication behaves on it.

Now II is a principal ideal (g(x))(g(x)) for some polynomial g(x)=b 0++b nx ng(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 ξ= n0a nx n nkx n\xi = \sum_{n \geq 0} a_n x^n \in \prod_n k \cdot x^n vanishes on II, in the sense that

p(x), na nx n=0\langle p(x), \sum_n a_n x^n \rangle = 0

for all p(x)(g(x))p(x) \in (g(x)), where the pairing ,:k[x]× nkx nk\langle , \rangle \colon k[x] \times \prod_n k \cdot x^n \to k is the canonical one:

x j, na nx n=a j.\langle x^j, \sum_n a_n x^n \rangle = a_j.

Taking the case p(x)=x jg(x)p(x) = x^j g(x), we obtain recurrence relations

b 0x j+b nx n+j, na nx n=b 0a j+b 1a j+1++b na n+j=0,\langle b_0 x^j + \ldots b_n x^{n+j}, \sum_n a_n x^n \rangle = b_0 a_j + b_1 a_{j+1} + \ldots + b_n a_{n+j} = 0,

one for every jj. We may pick a 0,a 1,,a n1a_0, a_1, \ldots, a_{n-1} at random, but from there on the remainder of the sequence a n,a n+1,a_n, a_{n+1}, \ldots is determined by the recurrence relations.

Now let g rev(x)g^{rev}(x) be the polynomial obtained by reversing the order of the coefficients b 0,,b nb_0, \ldots, b_n of gg,

g rev(x)=b n++b 0x n,g^{rev}(x) = b_n + \ldots + b_0 x^n,

and consider the problem of multiplicatively inverting g rev(x)g^{rev}(x) (which we may do since the constant coefficient of g revg^{rev} is non-zero). If ja jx j\sum_j a_j x^j is the reciprocal, we have

1=g rev(x) ja jx j=(b n++b 0x n)( ja jx j)1 = g^{rev}(x)\sum_j a_j x^j = (b_n + \ldots + b_0 x^n)(\sum_j a_j x^j)

which yields a second set of recurrence relations by examining coefficients of powers of xx. We get the first set of recurrence relations by restricting attention to the coefficients of x nx^n, x n+1x^{n+1}, etc.

Thus the power series of g rev(x) 1g^{rev}(x)^{-1} yields a ξ\xi which vanishes on I=(g(x))I = (g(x)), and similarly the power series of

xg rev(x),,x n1g rev(x)\frac{x}{g^{rev}(x)}, \ldots, \frac{x^{n-1}}{g^{rev}(x)}

also vanish on II. They in fact form a basis for the subspace (k[x]/(g(x))) * nkx n(k[x]/(g(x)))^\ast \hookrightarrow \prod_n k \cdot x^n.

We conclude that an arbitrary element of T(k) T(k)^\circ is of the form p(x)h(x)\frac{p(x)}{h(x)} for some hh with non-zero constant coefficient:


As subspaces of the space of formal power series nkx n\prod_n k \cdot x^n, there is an identification

T(k) =k[x] (x)T(k)^\circ = k[x]_{(x)}

where the right side is the localization of k[x]k[x] at the prime ideal (x)(x), whose elements are considered as formal power series.

It remains to identify the coalgebra structure on k[x] (x)k[x]_{(x)}. This extends the standard coalgebra structure on the tensor algebra k[x]k[x], where the comultiplication is given by deconcatenation:

δ(x N)= m+n=Nx mx n.\delta(x^N) = \sum_{m+n = N} x^m \otimes x^n.

We may as well focus on a typical subcoalgebra (k[x]/(g(x))) *(k[x]/(g(x)))^\ast, with basis elements x ig rev(x)\frac{x^i}{g^{rev}(x)}. By duality, the definition of δ(x ig rev(x))\delta(\frac{x^i}{g^{rev}(x)}) can be extracted from the multiplication table for residue classes x j(modg)(x)x^j \pmod g(x).

The calculations work out cleanest if we assume if kk is algebraically closed. For in that case, we can take advantage of partial fraction decompositions of the form

x ig rev(x)= kA k(1r kx) n k\frac{x^i}{g^{rev}(x)} = \sum_k \frac{A_k}{(1- r_k x)^{n_k}}

so that it suffices to give δ(1(1rx) n)\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)=(rx) n(1rx) n+1e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}} for n0n \geq 0. The power series expansion is

e n(r)= k0(kn)r kx ke_n(r) = \sum_{k \geq 0} \binom{k}{n} r^k x^k

and we formally calculate

δ(e n(r)) = k(kn)r kδ(x k) = k(kn)r k p+q=kx px q = p,q0(p+qn)r px pr qx q = p,q i+j=n(pi)(qj)r px pr qx q = i+j=n( p(pi)r px p)( q(qj)r qx q) = i+j=ne i(r)e j(r)\array{ \delta(e_n(r)) & = & \sum_k \binom{k}{n} r^k \delta(x^k) \\ & = & \sum_k \binom{k}{n} r^k \sum_{p+q=k} x^p \otimes x^q \\ & = & \sum_{p, q \geq 0} \binom{p+q}{n} r^p x^p \otimes r^q x^q \\ & = & \sum_{p, q} \sum_{i+j=n} \binom{p}{i} \binom{q}{j} r^p x^p \otimes r^q x^q \\ & = & \sum_{i+j=n} (\sum_p \binom{p}{i} r^p x^p) \otimes (\sum_q \binom{q}{j} r^q x^q) \\ & = & \sum_{i+j=n} e_i(r) \otimes e_j(r) }

which exhibits comultiplication as deconcatenation, familiar from the usual coalgebra structure on k[x]k[x]. In particular, we have δ(e 0(r))=e 0(r)e 0(r)\delta(e_0(r)) = e_0(r) \otimes e_0(r): the elements e 0(r)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)k[x]_{(x)} is indeed adjoint to multiplication on k[x]k[x]:

x px q,e n(r) = x p+q, k(kn)r kx k = (p+qn)r p+q = i+j=n(pi)(qj)r pr q = i+j=nx p,e i(r)x q,e j(r) = i+j=nx px q,e i(r)e j(r) = x px q,δ(e n(r)).\array{ \langle x^p x^q, e_n(r) \rangle & = & \langle x^{p+q}, \sum_k \binom{k}{n} r^k x^k \rangle \\ & = & \binom{p+q}{n} r^{p+q} \\ & = & \sum_{i+j=n} \binom{p}{i} \binom{q}{j} r^p r^q \\ & = & \sum_{i+j=n} \langle x^p, e_i(r) \rangle \langle x^q, e_j(r) \rangle \\ & = & \sum_{i+j=n} \langle x^p \otimes x^q, e_i(r) \otimes e_j(r) \rangle \\ & = & \langle x^p \otimes x^q, \delta(e_n(r)) \rangle. }

An analogous calculation shows that the counit ε:k[x] (x)k\varepsilon \colon k[x]_{(x)} \to k is defined by ε(e n(r))=δ 0n\varepsilon(e_n(r)) = \delta_{0 n} (Kronecker delta).

We summarize the calculations above as follows.


Let kk be an algebraically closed field. For each nonzero scalar rkr \in k, let L(r)L(r) denote the kk-linear span of the elements e n(r)=(rx) n(1rx) n+1e_n(r) = \frac{(r x)^n}{(1 - r x)^{n+1}} in k[x] (x)k[x]_{(x)}, with coalgebra structure given in each case by

δ(e n(r))= i+j=ne i(r)e j(r),ε(e n(r))=δ 0n\delta(e_n(r)) = \sum_{i+j=n} e_i(r) \otimes e_j(r), \qquad \varepsilon(e_n(r)) = \delta_{0 n}

Let L(0)=k[x]L(0) = k[x] as a subspace of k[x] (x)k[x]_{(x)}. The cofree coalgebra over a 1-dimensional space kk, identified with k[x] (x)k[x]_{(x)}, decomposes as a direct sum of coalgebras

k[x] (x) rkL(r)k[x]_{(x)} \cong \bigoplus_{r \in k} L(r)

according to the partial fraction decomposition of rational functions. Each of the coalgebras L(r)L(r) is isomorphic to k[x]k[x] with its standard coalgebra structure.


  • Walter Michaelis, Coassociative Coalgebras, Handbook of Algebra Volume 3, Elsevier (2003).

See also

  • Tom Fox, The construction of cofree coalgebras. JPAA 84 (1993) 191-198. texfile here

but note that this contains an error which is corrected in

Revised on August 29, 2015 23:13:20 by Todd Trimble (