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Aug 19, 2016 at 8:11 comment added Tim Porter Further thought, Whitehead's original papers on crossed complexes introduced the $\Gamma_n$ functors and in the 1991 paper of Carrasco and Cegarra they prove that the vanishing of the $\Gamma_n(X)$ is equivalent to the homotopy type of $X$ being completely modelled by its crossed complex. This can also be derived from results in the work of Ronnie Brown and Phil Higgins, as well as being generalised by Baues. The condition of $M\cap D=1$ and its interpretation still could do with exploration however.
Sep 25, 2014 at 12:51 comment added Omar Antolín-Camarena $Q=\Omega^\infty\Sigma^\infty$ so that $QS^2 = \mathrm{colim}_{n\to\infty} \Omega^n S^{n+2}$. I'd like to know about algebraic models for its 3-type too. (I'm sadly ignorant about that sort of thing, all I know is that this one is not modeled by a strict 3-groupoid.)
Sep 25, 2014 at 6:12 comment added Tim Porter Thanks for the clarification. Can you add what is $Q$ to help `the reader'? Has anyone calculated the algebraic models (crossed square, or whatever) for this 3-type?
Sep 24, 2014 at 15:16 comment added Omar Antolín-Camarena Minor correction: having vanishing Whitehead products does not imply a space is a product of Eilenberg-MacLane spaces. Consider the space $X = P_3 Q S^2$. Since it is a 3-type, the only Whitehead product to worry about is $\pi_2 \times \pi_2 \to \pi_3$, which vanishes since it agrees with the one for $QS^2$, which is an infinite loop space. On the other hand $X$ is not equivalent to $K(\pi_2X,2) \times K(\pi_3X,3)$ because the operation $\pi_2 \to \pi_3$ given by precomposing with the generator of $\pi_3(S^2)$ is non-zero on $X$ (but vanishes for the product of Eilenberg-MacLane spaces).
Sep 1, 2014 at 16:27 history edited Tim Porter CC BY-SA 3.0
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Sep 1, 2014 at 12:03 vote accept Tomer Schlank
Sep 1, 2014 at 6:17 history answered Tim Porter CC BY-SA 3.0