Edit: The description above of the cofree coalgebra is incorrect. I learned the correct version from Alex Chirvasitu. The description is as follows. Let $V$ be a vector space, and write $\mathcal T(V)$ for the tensor algebra of $V$, i.e. for the free associative algebra generated by $V$. Then the cofree coassociative algebra cogenerated by $V$ is constructed as follows. First, construct $\mathcal T(V^\ast)$, and second construct its finite dual $\mathcal T(V^\ast)^\circ$, which is the direct limit of duals to finite-dimensional quotients of $\mathcal T(V^\ast)$. There is a natural inclusion $\mathcal T(V^\ast)^\circ \hookrightarrow \mathcal T(V^\ast)^\ast$, and a natural map $\mathcal T(V^\ast)^\ast \to V^{\ast\ast}$ dual to the inclusion $V^\ast \to \mathcal T(V^\ast)$. Finally, construct $\operatorname{Cofree}(V) = \operatorname{Cofree}(V)$ as the union of all subcoalgebras of $\mathcal T(V^\ast)^\circ$ that map to $V \times_{V^{\ast\ast}} subseteq V^{\ast\ast}$ under the map $\mathcal T(V^\ast)^\circ \hookrightarrow \mathcal T(V^\ast)^\circ$T(V^\ast)^\ast \to V^{\ast\ast}$. Details are in section 6.4 (and specifically 6.4.2) of the book Hopf Algebras by Moss E. Sweedler. 4 corrected the description of the cofree coalgebra. Edit: It seems that some completion The description above of the tensor cofree coalgebra is incorrect. I learned the correct version from Alex Chirvasitu. The description is as follows. Let$V$be a vector space, and write$\mathcal T(V)$for the tensor algebra of$V$, i.e. for the free associative algebra generated by$V$. Then the cofree coassociative algebra cogenerated by$V$is neededconstructed as follows. This bothers me First, construct$\mathcal T(V^\ast)$, and second construct its finite dual$\mathcal T(V^\ast)^\circ$, which is the direct limit of duals to finite-dimensional quotients of$\mathcal T(V^\ast)$. There is a bitnatural inclusion$\mathcal T(V^\ast)^\circ \hookrightarrow \mathcal T(V^\ast)^\ast$, and a natural map$\mathcal T(V^\ast)^\ast \to V^{\ast\ast}$dual to the inclusion$V^\ast \to \mathcal T(V^\ast)$. Finally, construct$\operatorname{Cofree}(V) = V \times_{V^{\ast\ast}} \mathcal T(V^\ast)^\circ$. In any case,$\operatorname{Cofree}(V)$is something like the coalgebra of "finitely supported distributions on$V$" (or, anyway, that's is how to think of it in the cocomutative version). For example, when$V = \mathbb k$is one-dimensional, and$\mathbb k = \bar{\mathbb k}$is algebraically closed, then$\operatorname{Cofree}(V) = \bigoplus_{\kappa \in \mathbb k} \mathcal T(\mathbb k)$. I should emphasize that now when I write$\mathcal T(\mathbb k)$, in characteristic non-zero I do not mean to give it the Hopf algebra structure. Rather,$\mathcal T(\mathbb k)$has a basis$\lbrace x^{(n)}\rbrace$, and the comultiplication is$x^{(n)} \mapsto \sum x^{(k)} \otimes x^{(n-k)}$. Identifying$x^{(n)} = x^n/n!$, this is the comultiplication on the "divided power" algebra. It's reasonable to think of the$\kappa$th summand as consisting of (divided power) polynomials times$\exp(\kappa x)$, but maybe better to think of it as the algebra of descendants of$\delta(x - \kappa)$but this is just some Fourier duality. In the non-algebraically-closed case, there are also summands corresponding to other closed points in the affine line. end edit 3 hrm, I think something is wrong There cannot be a "free coalgebra" functor, at least in what I think is the standard usage. Namely, suppose that "orange" is a type of algebraic object, for which there is a natural "forgetful" functor from "orange" objects to "blue" objects. Then the "free orange" functor from Blue to Orange is the left adjoint, if it exists, to the forgetful map from Orange to Blue. Suppose that the forgetful map from coalgebras to vector spaces had a left adjoint; then it would itself be a right adjoint, and so would preserve products. Now the product in the category of coalgebras is something huge — think about the coproduct in the category of algebras, which is some sort of free product — and it's clear that the forgetful map does not preserve products. On the other hand, the coproduct in the category of coalgebras is given by the direct sum of underlying vector spaces, and so the forgetful map does preserve coproducts. This suggests that it may itself be a left adjoint; i.e. it may have a right adjoint from vector spaces to coalgebras, which should be called the "cofree coalgebra" on a vector space. Let me assume axiom of choice, so that I can present the construction in terms of a basis. Then I believe that the cofree coalgebra on the vector space with basis$L$(for "letters") is the graded vector space whose basis consists of all words in$L$, with the comultiplication given by$\Delta(w) = \sum_{a,b| ab = w} a \otimes b$, where$a,b,w$are words in$L$. I.e. the cofree coalgebra has the same underlying vector space as the free algebra, with the dual multiplication. It's clear that for finite-dimensional vector spaces, the cofree coalgebra on a vector space is (canonically isomorphic to) the graded dual of the free algebra on the dual vector space. Anyway, this is clearly a coalgebra, and the map to the vector space is zero on all words that are not singletons and identity on the singletons. I haven't checked the universal property, though. Edit: It seems that some completion of the tensor space is needed. This bothers me a bit. I should mention that in my mind the largest difference between algebras and coalgebras (by which I mean, and I assume you mean also, "associative unital algebras in Vect" and "coassociative counital coalgebras in Vect", respectively) is one of finiteness. You hinted at the difference in your answer: if$A$is a (coassociative counital) coalgebra (in Vect), then it is a colimit (sum) of its finite-dimensional subcoalgebras, and moreover if$X$is any$A$-comodule, then$X$is a colimit of its finite-dimensional sub-A-comodules. This is absolutely not true for algebras. It's just not the case that every algebra is a limit of its finite-dimensional quotient algebras. A good example is any field of infinite-dimension. It follows from the finiteness of the corepresentation theory that a coalgebra can be reconstructed from its category of finite-dimensional corepresentations. Let$A$be a coalgebra, $\text{f.d.comod}_A$ its category of finite-dimensional right comodules, and $F : \text{f.d.comod}_A \to \text{f.d.Vect}$ the obvious forgetful map. Then there is a coalgebra $\operatorname{End}^\vee(F)$, defined as some natural coequalizer in the same way that the algebra of natural transformations $F\to F$ is defines as some equalizer, and there is a canonical coalgebra isomorphism $A \cong \operatorname{End}^\vee(F)$. (Proof: see André Joyal and Ross Street, An introduction to Tannaka duality and quantum groups, Category theory (Como, 1990), Lecture Notes in Math., vol. 1488, Springer, Berlin, 1991, pp. 413–492. MR1173027 (93f:18015).) For an algebra, on the other hand, knowing its finite-dimensional representation theory is not nearly enough to determine the algebra. Again, the example is of an infinite-dimensional field (e.g. the field of rational functions). On the other hand, it is true that knowing the full representation theory of an algebra determines the algebra. Namely, if$A$is an (associcative, unital) algebra (in Vect), $\text{mod}_A$ its category of all right modules, and $F: \text{mod}_A \to \text{Vect}$ the forgetful map, then there is a canonical isomorphism $A \cong \operatorname{End}(F)$. (Proof:$F$has a left adjoint,$V \mapsto V\otimes A$. But$V \mapsto V\otimes \operatorname{End}(F)$is also left-adjoint to$F$. The algebra structure comes from the adjunction: the $\text{mod}_A$ map$A\otimes A \to A$corresponds to the vector space map$\operatorname{id}: A\to A$.) ((Note that you don't actually need the full representation theory, which probably doesn't exist foundationally, but you do need modules at least as large as$A\$.))