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I am reading the definition of Bialgebras over a field $\mathbb{K}$. The definition is the following:

A bialgebra over a field $\mathbb{K}$ is a vector space $B$ over $\mathbb{K}$ equipped with $\mathbb{K}$-linear maps (multiplication) $\nabla : B \otimes B \to B$, (unit) $\eta: \mathbb{K} \to B$, (comultiplication) $\Delta: B \to B \otimes B$, and (counit) $\varepsilon: B \to \mathbb{K}$ such that:

  1. $(B,\nabla,\eta)$ is a unital associative algebra.

  2. $(B,\Delta,\varepsilon)$ is a (counital, coassociative) coalgebra.

along with the following four compatibility conditions:

  1. Relation between multiplication $\nabla$ and comultiplication $\Delta$:

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where $\mu: B \otimes B \to B$ is the linear map defined by $\mu(x \otimes y) = y \otimes x$ for all $x, y \in B$.

  1. Relation between multiplication $\nabla$ and counit $\varepsilon$ and comultiplication $\Delta$ and unit $\eta$:

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  1. Relation between unit $\eta$ and counit $\varepsilon$:

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The four commutative diagrams can be read either as "comultiplication and counit are homomorphisms of algebras" or, equivalently, "multiplication and unit are homomorphisms of coalgebras".


My Question: How can we conclude the statement "comultiplication and counit are homomorphisms of algebras" from the above diagrams?

Saikat
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1 Answers1

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The first diagram says for $f,g\in B$: $$\Delta\circ\nabla(f\otimes g)=(\nabla\otimes\nabla)\circ(\operatorname{id}\otimes\tau\otimes\operatorname{id})\circ (\Delta(f)\otimes (\Delta g)).$$ Develop both sides and see what pops out.

The other one is easier. The second diagram says for $f,\,g\in B$: $$\varepsilon(\nabla(f\otimes g))=(\varepsilon\otimes\varepsilon)(f\otimes g).$$ Develop this.

JP McCarthy
  • 8,209