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

  • 1
    $\begingroup$ Universal in the homotopy category, I presume? $\endgroup$ Jan 8, 2015 at 22:24
  • 5
    $\begingroup$ Okay, so I'm pretty sure that even with the hypothesis on $\pi_0$, $\pi_1$ already presents a problem; in particular it should be possible to find a nonabelian group $G$ such that $BG$ does not admit a simplification. (Note that $B (G/[G, G])$ is not a simplification; $G$ can be perfect and $BG$ can have nontrivial higher cohomology.) $\endgroup$ Jan 8, 2015 at 22:57
  • 1
    $\begingroup$ Oh, hmm, never mind. If $G$ is perfect then I think the plus construction applied to $BG$ might be a simplification...? $\endgroup$ Jan 8, 2015 at 23:15
  • 4
    $\begingroup$ Eric Wofsey's excellent example reminds me of Mislin's example showing that localizations in the homotopy category need not exist. But one can ask about homotopy localizations (using mapping spaces rather than sets of homotopy classes). Is there any map $f$ such that $f$-local spaces are precisely the simple spaces? $\endgroup$ Jan 8, 2015 at 23:33
  • 4
    $\begingroup$ This illustrates something interesting. A space is simple if and only if every map $S^1 \vee S^n \to X$ extends to a map $S^1 \times S^n \to X$, and so you might expect to construct such a "simplification" using a Bousfield localization which inverts such maps. But Eric's answer indicates that this localization must do something unexpected. $\endgroup$ Jan 8, 2015 at 23:33

3 Answers 3


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.

  • 2
    $\begingroup$ I'm being stupid, I know, but could you explain the last sentence further? $\endgroup$
    – Mark Grant
    Jan 14, 2015 at 13:47
  • 3
    $\begingroup$ The universal property says that $f$ induces a bijection between homotopy classes of maps $S^1\vee S^1\to K(\mathbb{Z},2)$ and homotopy classes of maps $X\to K(\mathbb{Z},2)$. Explicitly, the trivial map $S^1\vee S^1\to K(\mathbb{Z},2)$ can be factored through $X$, but the factorization is not unique. $\endgroup$ Jan 14, 2015 at 18:42

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.

  • 3
    $\begingroup$ Thanks for adding your answer. I think this is a great example. $\endgroup$ Feb 3, 2015 at 13:49

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 $H_\ast$).

  • 3
    $\begingroup$ I don't believe the statement that "every local coefficient system on $X^s$ is trivial". For instance, if $X=K(A,1)$ for $A$ an abelian group, then $X^s=X$ and local coefficient systems on $X^s$ are essentially $\mathbb{Z}[A]$-modules. Of course, in this case $X\to X^s$ is surely acyclic! $\endgroup$
    – Mark Grant
    Feb 28, 2018 at 9:32
  • $\begingroup$ You're right of course. If that pithy solution doesn't work, I'm not sure how valuable it is to pursue this question further, but I've done so anyway :) $\endgroup$
    – Tim Campion
    Feb 28, 2018 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.