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I am reading an very interesting paper ''The infinitesimal Hopf algebra and the poset of planar forests'' https://arxiv.org/pdf/0802.0442.pdf written by Pro. Foissy. In his paper, on page 4, I don't understand why $\epsilon(a^{(1)})\epsilon(a^{(2)}b)=\epsilon(ab)$? Here we use the sweedle notation $\Delta(a)=a^{(1)}\otimes a^{(2)}$.

By the counicity I accept that $\epsilon(a^{(1)})\epsilon(a^{(2)})=\epsilon(a)$. Hint: under the view of infinitesimal Hopf algebra, the multiplication $m$ is not a coalgebra morphism, and the coproduct $\Delta$ is not an algebra morphism. It satisfies $\Delta (ab)=\Delta(a)(1\otimes b)+(a\otimes1)\Delta(b)-a\otimes b$. Then we can't use $\epsilon\circ m=\epsilon \otimes \epsilon $.

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  • $\begingroup$ So the counit is explicitly not an algebra morphism? $\endgroup$
    – Joppy
    Apr 13 '18 at 7:33
  • $\begingroup$ @Joppy, I think so $\endgroup$
    – Daisy
    Apr 13 '18 at 8:30
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By the property of the counit, we have $\varepsilon(a^{(1)})a^{(2)} = ((\varepsilon\otimes 1)\circ \Delta)(a)= a$ for all $a\in A$.
As $\varepsilon$ is also $K$-linear, we obtain $\varepsilon(a^{(1)}) \varepsilon(a^{(2)}b) = \varepsilon\bigl(\varepsilon(a^{(1)})a^{(2)}b\bigr) = \varepsilon(ab)$ for all $a,b\in A$.

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  • $\begingroup$ Thank you very much. I ignore the $K$-linear. $\endgroup$
    – Daisy
    Apr 13 '18 at 14:14

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