Is there a "simplification" functor in algebraic topology?

Question

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 $\pi_2(\Omega \Sigma \mathbb{R}\mathrm{P}^2)$.)

Answer 1

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.

Answer 2

So I just wanted to point out an aside to this, which I started to mention in the comments above.

A space $X$ is simple if, and only if, any map $S^1 \vee S^n \to X$ extends to a map $S^1 \times S^n \to X$. You can see this by looking at the attaching map. The map $S^1 \vee S^n \to S^1 \times S^n$ attaches exactly one new cell. If $g$ is the generator of $\pi_1 S^1$ and $h$ is the generator of $\pi_n S^n$, the attaching map is the commutator $[g,h]$ if $n=1$, and the element $g\cdot h - h$ if $n > 1$. By definition $X$ is simple if and only if such elements are always trivial.

Let's say a space $X$ is very simple if, for any n, the map $Map(S^1 \times S^n,X) \to Map(S^1 \vee S^n,X)$ is an equivalence. There is a "very simplification" functor on spaces: for very general reasons there is a left Bousfield localization functor on the category of spaces which takes a space $X$ to a very simple space, effectively forcing the maps $\{S^1 \vee S^n \to S^1 \times S^n\}_{n \geq 1}$ to become equivalences. This is a "standard" technique for trying to construct a simplification functor like you describe.

It turns out that a space $X$ is very simple if and only if, for any basepoint $x$:

• the fundamental group $\pi_1(X,x)$ is abelian, and

• all the groups $\pi_k(X,x)$ are trivial for $k > 1$.

The sufficiency of this is a consequence of obstruction theory, and the first condition is necessary. Let me show that the second is also necessary. If $X$ is very simple and $x \in X$, we take the pullback of the diagram $$ \{x\} \to Map(S^1 \vee S^n, X) \xleftarrow{\sim} Map(S^1 \times S^n, X) $$ where the right-hand map is a Serre fibration and an equivalence. The pullback, which is equivalent to a point, is the set of maps $S^1 \times S^n \to X$ which restrict to the constant map $S^1 \vee S^n \to \{x\} \subset X$: this is the same as the space of based maps $S^1 \wedge S^n \to X$, alias $\Omega^{n+1} X$. Since this is equivalent to a point, all the higher homotopy groups of $X$ are trivial.

(Note that this proof shows that we only need $X$ to satisfy this condition for $n=1$, or for the maps $Map(S^1 \times S^n,X) \to Map(S^1 \vee S^n,X)$ to have connected homotopy fibers, for this result to hold.)

Thanks for asking the question, because it underscored that I don't understand Bousfield localizations as well as I'd thought.

Answer 3

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:

• $X$ such that $\pi_1(X)$ is perfect

• $X$ such that $X^+$ is an $H$-space (e.g. all the cases used in algebraic $K$-theory)

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 $H_\ast$).