I have just learned about the construction of the Eilenberg-Maclane space $K(A,n)$ ($n \geq 1$). I am using Hatcher's text. He briefly describes the construction on p.365. I will recap the construction here:
Choose a presentation $A = \langle a_{\alpha} | r_{\beta} \rangle$ (which exists since every group is a quotient of a free group, so the $a_{\alpha}$'s can be taken to be the generators of this free group with the $r_{\beta}$'s the generators of the kernel of the map from the free group to $A$). Define the CW complex $X_{n+1} = (\bigvee_{\alpha} S_{\alpha}^n) \cup_{\beta} e_{\beta}^{n+1}$, where the copies of $S^n$ are in bijection with the generators $a_{\alpha}$ of $A$ and the $(n+1)$-cells are glued corresponding to the relations $r_{\beta}$ of $A$. Then, $\pi_n(X_{n+1}) \cong A$, while $\pi_i(X_{n+1}) = 0$ for $i < n$ (which one can see by applying Proposition 1.26 of Hatcher's text in the n = 1 case, and using the Hurewicz Theorem for the case $n \geq 2$). Next, we attach $(n+2)$-cells to $X_{n+1}$ using cellular maps $S^{n+1} \rightarrow X^{n+1}$ that generate $\pi_{n+1}(X_{n+1})$ to form a space $X_{n+2}$ with $\pi_{n+1}(X_{n+2})$ trivial and $\pi_i(X_{n+2}) \cong \pi_i(X_{n+1})$ for $i \leq n$. Similarly, we attach $(n+3)$-cells to $X_{n+2}$ to form a space $X_{n+3}$ with $\pi_{n+2}(X_{n+3})$ trivial and $\pi_i(X_{n+3}) \cong \pi_i(X_{n+2})$ for $i \leq n+1$. We continue this process to obtain the desired CW complex $X = X_{\infty}$.
I have two questions about this construction:
- How can we see that $\pi_{n+1}(X_{n+2})$ is trivial? Are we choosing a presentation for the group $\pi_{n+1}(X_{n+1})$ and seeing how this group is affected once we attach $(n+2)$-cells to $X_{n+1}$ via the generators of this group? What's the importance of those maps being cellular?
- My instructor says that any construction he knows of $K(A,n)$ requires a colimit. I don't see one involved here. Does the colimit come in just in the last line above, to make the construction $X = X_{\infty}$ rigorous?
Thanks!