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I realise that this question might be rather basic but however I was unable to find the answer in any textbook nor manage to figure out the answer. The question is the following: given two Hopf algebras $H_1,H_2$ is there a canonical way to turn the direct sum $H_1 \oplus H_2$ into a Hopf algebra? I have a problem already at the level of bialgebras: the counit is always a algebra morphism (linear and multiplicative) but the set of all linear multiplicative maps $\omega: H_1 \oplus H_2 \to \mathbb{C}$ is a sum of such sets corresponding to $H_1$ nad $H_2$. So I don't see how to define (in a canonical way) the counit map $\varepsilon_{H_1 \oplus H_2}$.

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The answer is "no". In the commutative case, this would be asking for the disjoint union of two group schemes to be a group scheme. Even more concretely (say, working with finite dimensional commutative Hopf algebras over $\mathbb{C}$), there is no natural way to make the disjoint union of two finite groups into a finite group. The algebras of functions on the groups would be the Hopf algebras in question. Your difficulty with the counit amounts to the problem of making a canonical pointed set from the disjoint union of two pointed sets.

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Nice question and nice answer. Maybe groupoids and Hopf algebroids can come in to solve the problem? – Fernando Muro Dec 23 '13 at 8:36
@FernandoMuro That is an excellent suggestion. If we expand our scope to Hopf algebroids, we get a well-behaved direct sum, and in the commutative case, this yields the disjoint union of groupoids of affine schemes. – S. Carnahan Dec 23 '13 at 9:50

The answer to your question appears to be given in Comm. Math. Phys.(1999) 38(2), 519-524 where it's mentioned that because $(H_1\oplus H_2)\otimes (H_1\oplus H_2)\neq (H_1\otimes H_1)\oplus (H_2\otimes H_2)$ we cannot in general construct a Hopf algebra on $H_1\oplus H_2$ directly from the Hopf algebra structures on $H_1$ and $H_2$. However, they do define a separate direct sum of Hopf algebras which can be regarded as a Hopf algebra.

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