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Suppose that $\mathcal{O}$ is a differential graded operad over a field and that $H(\mathcal{O})$ (i.e. taking the arity wise homology) is an operad, too. (If possible, I would avoid to restrict to symmetric operads)

Is the natural projection $\pi : \mathcal{O} \to H(\mathcal{O})$ necessarily a morpism of operads?

Mark Neuhaus
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    Taking the aritywise homology always gives an operad. – Qiaochu Yuan Sep 25 '15 at 22:17
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    Just a comment: over a field, it's true that for any chain complex $C_$ there always exists a splitting $C_ = H_(C) \oplus V \oplus V[1]$ where the differential sends $v \in V$ to $v \in V[1]$ (zero elsewhere), and you can use that to project $C_$ onto $H_(C)$. But there is no way to choose that splitting canonically, and usually no way to choose it in each arity to get a morphism of operads $\mathcal{O} \to H(\mathcal{O})$. Over a general ring, the splitting that I've mentioned doesn't even necessarily exist (and there are probably chain complexes with no projection $C_ \to H_*(C)$). – Najib Idrissi Oct 01 '15 at 19:26
  • Ah ok. Good to know. Thanks! This is in particular useful since many operads are relatively simple vector spaces, arity wise. – Bobby Oct 02 '15 at 20:28

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There isn't any such natural projection.

Qiaochu Yuan
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  • What? Project first onto the kernel of the differential in homological degree $k$. This is a linear map. Then take the quotient with the image of the differential in degree $k+1$. What is the problem with this map? – Mark Neuhaus Sep 26 '15 at 05:36
  • @Mark: there's no natural way to project onto the kernel of the differential. – Qiaochu Yuan Sep 26 '15 at 06:16
  • Ah ok. I see. I would delete the question, but it's not possible anymore. – Mark Neuhaus Sep 26 '15 at 06:37