# $\pi_n(X^n)$ free Abelian?

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

$$f: (S^n, s_0) \rightarrow (X,x_0)$$

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?

• 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. – Joe Johnson 126 Mar 3 '15 at 19:39
• 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 $n$. – 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.

• 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! – Kevin Ye Mar 4 '15 at 23:07
• @KevinYe When I thought about it more, it is only the $n+1$ cells. I was being cautious by adding $n+2$. – Joe Johnson 126 Mar 5 '15 at 15:22
• Just wondering that if I am missing something, why is $\pi_{n}(X^{n-1})=0$? – Rainbow Mar 6 '15 at 9:35
• @JoeJohnson126 Well, $S^{n-1}$ has no cells in dimension $n$ yet it has a non-trivial $\pi_{n}$. – Rainbow Mar 6 '15 at 19:18
• @Rainbow It is because we are dealing with Eilenberg-MacLane spaces, the map into $\pi_n (X^n)$ is trivial, by the construction. – Kevin Ye Mar 6 '15 at 19:20