The Heyneman-Radford theorem for monoidal categories (Q868850)

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The Heyneman-Radford theorem for monoidal categories
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    The Heyneman-Radford theorem for monoidal categories (English)
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    26 February 2007
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    Let \(E\) and \(C\) be coalgbras over a field, \(D\) the coradical of \(E\). Let \(f\) be a coalgebra morphism from \(E\) to \(C\) such that the restriction of \(f\) to \(D^D\) is injective. \(D^D\) (the wedge product of \(D\) with itself) is the set of elements in \(E\) whose coproduct is in \(E\otimes D\oplus D\otimes E\). The Heyneman-Radford theorem [\textit{R. G. Heyneman} and \textit{D. E. Radford}, J. Algebra 28, 215--246 (1974; Zbl 0291.16008)] concludes that \(f\) is then injective. The author of the paper under review generalizes this to the framework of a monoidal category \(\mathcal M\) which is cocomplete and coabelian (the latter means that \(\mathcal M\) is abelian and tensoring (on either side) with an object of \(\mathcal M\) gives an additive left exact functor) and which satisfies AB5. Let \(D\) and \(E\) be coalgebras in \({\mathcal M}, d\) a monomorphism from \(D\) to \(E\) which is a coalgebra morphism in \(\mathcal M\), and let \((L,p)\) be the cokernel of \(d\) in \(\mathcal M\). Let \((D^D, d_2)\) be the kernel of the comultiplication in \(E\) followed by \(p\otimes p\). \(d_2\) is a coalgebra morphism from \(D^D\) to \(E\). Let \(f\) be a coalgebra morphism from \(E\) to \(C\) such that such that \(fd_2\) from \(D^D\) to \(C\) is a monomorphism. The generalized Heyneman-Radford theorem to \(\mathcal M\) concludes that the coalgebra morphism \({\tilde fd}\) from \({\tilde D}\) to \(C\) is a monomorphism. Here \({\tilde D}\) is a direct limit of higher wedge products of \(D\), and \({\tilde d}\) is a certain coalgebra morphism from \({\tilde D}\) to \(E\). The original Heyneman-Radford theorem is obtained by taking \(\mathcal M\) to be the category of vector spaces over a field. The author remarks that the proof of the generalization is different from the proof of the classical one, and should be of interest even in the classical case.
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