For a finite dimensional $k$-algebra $A$, each $A^{\otimes n}, n \geq 0$ is a $k$-algebra ($A^0 = k $). Let $T= \prod\_{n \geq 0} A^{\otimes n}$. This is a $k$-alegbra with unit $(1,1,\dots)$ and multiplication is component-wise. Let $\Delta^{(n)} : A^{\otimes n} \to T \otimes T$ be the deconcatenation map
$$ \Delta^{(n)}(a\_1 \otimes \dots \otimes a\_n) = \sum\_{i=0}^n (a\_1 \otimes \dots \otimes a\_i) \otimes (a\_{i+1} \otimes \dots \otimes a\_n ). $$


I want to extend these $\Delta^{(n)}$ to a comultiplication $\Delta : T \to T \otimes T $. This does not seem to work in a straightforward way because if $t = ( t\_0, t\_1, \dots ) \in T, $ then $\sum\_n \Delta^{(n)}(t\_n)$ may not be a finite sum of pure tensors in $T \otimes T$ (I have not shown this sum can be infinite, but suspect it can be).



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> Is there a way to make $T$ into a Hopf algebra so that $\Delta(t) = \Delta^{(n)}(t)$ when $t\_i=0$ for $i \ne n$? If not, is there an algebra similar to $T$ where this does work?
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> Is there a standard way to complete the tensor product and instead get a map $\Delta$ from $T$ to the completion? Does this give rise to a genuine Hopf algebra or some generalization of Hopf alegbras?
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Hopf algebra structure on $\prod_n A^{\otimes n}$ for an algebra $A$
You can indeed complete the tensor product and get a good comultiplication, but it's not strictly speaking a Hopf algebra. An algebraic geometer would call it an affine formal group. If you think of the infinite product $T=\prod\_n A^{\otimes n}$ as a pro-object indexed by $\mathbb{N}$ with $T(n) = \prod\_{i=0}^n A^{\otimes i}$, then you can tensor $T$ with itself and get a pro-object indexed by $\mathbb{N} \times \mathbb{N}$. Take the limit and that's the completed tensor product you want. Alternatively, you can do it with topologized rings and topologically-complete the tensor product.


For the uncompleted tensor product, there is no comultiplication extending your rule. There is, however, a cofree Hopf algebra (as Anton points out, consult [Does the forgetful functor {Hopf Algebras}→{Algebras} have a right adjoint?](https://mathoverflow.net/questions/22659/does-the-forgetful-functor-hopf-algebrasalgebras-have-an-adjoint)). That's a sub-Hopf-algebra of the completed Hopf algebra $T$.


