Sorry to resurrect an old question, but I just wanted to expand here on an observation that Qiaochu made above (in the form of a question, but I think he was just playing *Jeopardy!*).

~~ Let $X$ be a connected space, and suppose that it has a simplification $X \to X^s$. Since Eilenberg-MacLane spaces are simple, the map $X \to X^s$ is a (co)homology equivalence in all degrees. Since every local coefficient system on $X^s$ is trivial, this implies that $X \to X^s$ is an acyclic map. It also must kill the perfect radical of $\pi_1(X)$. But the plus construction $X \to X^+$ is the unique acyclic map killing the perfect radical of $\pi_1(X)$. Therefore $X^s = X^+$.~~

If $X^+$ happens to be simple, then $X \to X^+$ is a simplification, simply because any map $X \to Y$ with $Y$ simple must kill the perfect radical (which in this case must coincide with the commutator) of $\pi_1(X)$, and $X \to X^+$ is the universal map which does this.

So among the spaces $X$ that *do* have a simplification are:

We can also conclude that if $X$ has a simplification $X \to X^s$, then it factors through the plus construction $X \to X^+ \to X^s$, and $X^s$ is also the simplification of $X^+$. So in asking which spaces $X$ admit a simplification, we can reduce to the case where $\pi_1(X)$ is *solvable* (at least under suitable finiteness assumptions).

**Warning:** From here on, this discussion grows increasingly aimless.

Because Eilenberg-MacLane spaces are simple, a simplification is a (co)homology isomorphism.

Assume $X^s$ exists, and write $G = \pi_1(X), G^s = \pi_1(X^s)$. If $G \to G^s$ is not surjective, then it lifts to some cover $\overline{X^s}$ of $X^s$. On homology, $X^s$ is a retract of $\overline{X^s}$, but both spaces have abelian fundamental group so it's also a retract on $\pi_1$. But the map on $\pi_1$ is injective, so it is an isomorphism. Hence $G \to G^s$ *is* surjective after all, and is precisely the quotient by the commutator subgroup, i.e. we have $G^s = G^{ab}$.

So the cofiber of $X \to X^s$ is acyclic and is also simply-connected, so is contractible. Since the 1-truncation functor preserves cofiber sequences, the map $BG \to BG^{ab}$ likewise has trivial cofiber. So $BG \to BG^{ab}$ is a homology isomorphism.

I'm not sure where to go from there, so instead, let's pick off a special case. Assume that $G$ is abelian, so $G = G^{ab}= G^s$. Then by comparing fiber sequences $\tilde X \to X \to BG$ and $\tilde X^s \to X^s \to BG$ (where tilde's denote universal covers), we see that $H_\ast(\tilde X^s) = H_\ast(\tilde X)_G$ and in addition, the map $H_\ast(\tilde X) \to H_\ast(\tilde X)_G$ is an equivalence on $G$-homology, for any constant coefficients. So for example, if $G$ is a finitely-generated torsion-free abelian group and $H_\ast(X)$ is finitely-generated in each degree over $G$, then I *think* we can conclude that $X$ was already at least weakly simple (i.e. has trivial $\pi_1$-action on $H_\ast$).

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