I have encountered a problem which states that denote $X$ as an Eilenberg-MacLane space $K(G,1)$ and is a CW complex, show that $\pi_n(X^n)$ is free Abelian for $n \geqslant 2$.

However, I think I have a conceptual misunderstanding in this problem. According to cellular approximation, the map

\begin{equation} f: (S^n, s_0) \rightarrow (X,x_0) \end{equation}

can be homotoped to a cellular map so that it maps $(S^n,s_0)$ into $X^n$. In this sense the $\pi_n(X^n)$ should be equal to $\pi_n(X)$, which is actually zero.

I would appreciate for clarification in this case. Moreover, how do we show that the homotopy group is free Abelian?

  • 3
    $\begingroup$ It is true that the map can be homotoped into $X^n$. But, it becomes trivial due to the $X^{n+1}$ and $X^{n+2}$ skeletons, which $X^n$ lacks. $\endgroup$ – Joe Johnson 126 Mar 3 '15 at 19:39
  • $\begingroup$ For the reason mentioned above, it's a standard result that the inclusion $X^n \to X$ induces an isomorphism on homotopy groups below dimension $n$ and a surjection in dimension $n$. $\endgroup$ – anomaly Mar 3 '15 at 20:41

Joe Johnson 126's answer is incorrect; see Is it true that $\pi_n(X^{n-1})=0$ for $n>1$ if $X$ is $K(G,1)$ space?. Indeed, the claimed statement is very much not true for all CW-complexes. For instance, if $X=S^3$, then $\pi_4(X^4)=\pi_4(S^3)=\mathbb{Z}/2$.

Here is a correct answer. First of all, to address the first question, all your argument shows is that the map $\pi_n(X^n)\to\pi_n(X)$ is surjective. You know that every map $S^n\to X$ can be homotoped to a map $S^n\to X^n$, but it might be that two such maps are homotopic in $X$ but not in $X^n$. A homotopy of such maps is a map $S^n\times I\to X$, which by cellular approximation can only be homotoped to $X^{n+1}$, since $S^n\times I$ is $(n+1)$-dimensional. So you do have an isomorphism $\pi_n(X^{n+1})\to \pi_n(X)$.

Let's now solve the original problem. Note that by the argument above, $\pi_k(X^n)\cong \pi_k(X)$ for all $k<n$, so since $X$ is a $K(G,1)$, $\pi_k(X^n)$ is trivial for $k<n$ except for $k=1$. Let $Y$ be the universal cover of $X^n$; then $Y$ is also an $n$-dimensional CW-complex, and now $\pi_k(Y)=0$ for all $k<n$. By Hurewicz, we have $\pi_n(Y)\cong H_n(Y)$. But $H_n$ of any $n$-dimensional CW-complex is free abelian (because computing cellular homology, there are no nontrivial $(n+1)$-chains and hence no nontrivial $n$-boundaries, so $H_n$ is just the group of $n$-cycles, which is a subgroup of the free abelian group of $n$-chains). Thus $\pi_n(Y)$ is free abelian, and hence so is $\pi_n(X^n)\cong \pi_n(Y)$.


This is true of any CW-complex. We have an inclusion $X^{n-1}\hookrightarrow X^{n}$. The long exact sequence of the pair gives $$ 0\rightarrow \pi_{n}(X^{n})\rightarrow \pi_{n}(X^{n},X^{n-1})\rightarrow \pi_{n-1}(X^{n-1})\rightarrow \cdots $$ Since $X^n/X^{n-1}$ is a wedge of spheres, $\pi_n(X^n,X^{n-1})$ is free abelian. Then $\pi_n(X^n)$ is a subgroup of a free abelian group, hence free abelian.

EDIT: To address the question in the comments below: The inclusion $X^n\hookrightarrow X$ induces an isomorphism on homotopy groups for $k<n$ and a surjection at $k=n$. This is due to the same long exact sequence as above $$ \cdots\to \pi_{k+1}(X,X^n)\to \pi_k(X^n)\to \pi_k(X)\to \pi_k(X,X^n)\to \pi_{k-1}(X^n)\to \cdots. $$ If $k\leq n$ then cellular approximation tells us that all maps $S^k\to X$ can be homotoped into $X^n$. Thus $\pi_k(X,X^n)=0$ for $k\leq n$. Looking at the sequence above one sees that $\pi_k(X^n)\cong \pi_k(X)$ for $k<n$ and $\pi_n(X^n)$ surjects onto $\pi_n(X)$. Similarly, $\pi_n(X^{n+1})\cong \pi_n(X)$. Thus $\pi_n(X)$ only depends on the the $n+1$-skeleton.

  • $\begingroup$ When I rethink this today, there is one confusion about the first comment. To which order will the cells affect $\pi_n(X)$? Is it only $n+1$ cells, or as you wrote down, $n+1$ and $n+2$ cells? How do I understand this? Also, thank you very much for the explanation! $\endgroup$ – Kevin Ye Mar 4 '15 at 23:07
  • $\begingroup$ @KevinYe When I thought about it more, it is only the $n+1$ cells. I was being cautious by adding $n+2$. $\endgroup$ – Joe Johnson 126 Mar 5 '15 at 15:22
  • 1
    $\begingroup$ Just wondering that if I am missing something, why is $\pi_{n}(X^{n-1})=0$? $\endgroup$ – Rainbow Mar 6 '15 at 9:35
  • $\begingroup$ @JoeJohnson126 Well, $S^{n-1}$ has no cells in dimension $n$ yet it has a non-trivial $\pi_{n}$. $\endgroup$ – Rainbow Mar 6 '15 at 19:18
  • $\begingroup$ @Rainbow It is because we are dealing with Eilenberg-MacLane spaces, the map into $\pi_n (X^n)$ is trivial, by the construction. $\endgroup$ – Kevin Ye Mar 6 '15 at 19:20

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.