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Mar 27, 2013 at 2:18 comment added Peter May 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.
Mar 25, 2013 at 19:32 comment added Bad English Thank you. What for spaces without h-group structure? Is a Hurewicz map to primitive elements surjective?
Mar 25, 2013 at 19:07 vote accept Bad English
Mar 24, 2013 at 22:50 history answered Peter May CC BY-SA 3.0