I heard this statement for loop spaces, but can't find proof for that or counter-example for following question. consider ordinary rational homologies, there are primitive elements with respect to coalgebra structure, it is almost obvious that homology classes released by spheres (with rational coefficient) are such elements, is it all?
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Yes. This is classical, maybe originally in Milnor and Moore's paper on Hopf algebras. For a recent exposition see for example "More concise algebraic topology" by Kate Ponto and myself. If $X$ is a connected $H$-space (say with finitely generated rational homology groups), then the rationalized Hurewicz homomorphism is a monomorphism with image the primitive elements of $H_*(X;\mathbf Q)$. The essential point is that the rationalization of $X$ is equivalent to a product of Eilenberg-MacLane spaces.
answered Mar 24, 2013 at 22:50
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$\begingroup$ Thank you. What for spaces without h-group structure? Is a Hurewicz map to primitive elements surjective? $\endgroup$ Commented Mar 25, 2013 at 19:32
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1$\begingroup$ Don't have time to think about a counterexample, but it is clear from how Sullivan rational homotopy theory works that they are plentiful. That theory gives a rational DGA A(X) for a nilpotent (or simply connected in the original) space X such that the cohomology of A(X) is the rational cohomology of X and the indecomposables of A(X) are dual to the rationalized homotopy groups of X. The cohomology of A(X) can have indecomposables that are decomposable in A(X). No reason the Hurewicz homomorphism should hit their duals. $\endgroup$ Commented Mar 27, 2013 at 2:18