Recall that a space (=CW complex) is called simple if it is connected, the fundamental group is abelian, and the fundamental group acts trivially on all higher homotopy groups. Call Simp(X) a simplification of X if it is universal for maps from X to a simple space. Does Simp(X) exist for any connected space X? This would give a higher analogue of abelianization of groups.


(A natural guess would be the loop suspension of X, but I'm pretty sure that doesn't work as the following example shows. Let X be $\mathbb{R}\mathrm{P}^2$, and note that the $2$-type of $X$ can be described completely by saying that $\pi\_1(X) = \mathbb{Z}/2$, $\pi\_1(X) = \mathbb{Z}$, the action is the nontrivial one sending $n \mapsto -n$, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z})$. There is a simple space $Y$ which is a $2$-type where $\pi\_1(X) = \mathbb{Z}/2$, $\pi\_2(X) = \mathbb{Z}/2$, the action is trivial, and the gluing is by the nontrivial element of $H^3(\mathbb{Z}/2,\mathbb{Z}/2)$. There's a natural map from X to Y since the nontrivial action on $\mathbb{Z}$ becomes trivial when you kill $2$. On the other hand, $\pi\_2(\mathbb{R}\mathrm{P}^2) = \mathbb{Z}$ maps to $\pi\_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2) = \mathbb{Z}/4$ sending $1 \mapsto 2$, and so the surjective map $\mathbb{Z} = \pi\_2(X) \rightarrow \pi\_2(Y) = \mathbb{Z}/2$ cannot factor through $\pi\_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2)$.)



Is there a "simplification" functor in algebraic topology?
The space $S^1\vee S^1$ does not have a simplification. Indeed, suppose $f:S^1\vee S^1\to X$ is a simplification and let $i:S^1\vee S^1\to S^1\times S^1$ be the standard inclusion. Then since $X$ is simple, the commutator of the generators of $\pi\_1(S^1\vee S^1)$ becomes nullhomotopic after composing with $f$, so $f$ extends over $i$ to a map $S^1\times S^1\to X$. But $S^1\times S^1$ is already simple, so $i$ must factor through $f$. Thus we have maps $S^1\times S^1\to X\to S^1\times S^1$, and the composition induces the identity on $\pi\_1$ and is thus homotopic to the identity. This implies that $H\_2(X)$ has $\mathbb{Z}$ as a direct summand, and thus $H^2(X)$ is nontrivial. But this is a contradiction since $H^2(S^1\vee S^1)=0$ and $K(\mathbb{Z},2)$ is simple.


