How to construct $K(\mathbb{Z}/5\mathbb{Z},1)$?

Question

From the Wikipedia article on Eilenberg-MacLane spaces:

A $K(G, n)$ can be constructed stage-by-stage, as a CW complex, starting with a wedge of n-spheres, one for each generator of the group $G$, and adding cells in (possibly infinite number of) higher dimensions so as to kill all extra homotopy.

I have a vague intuition of how this would work—for, say, $\mathbb{Z}/5\mathbb{Z}$—but I haven't been able to locate a step-by-step description of such a construction. I know the fact that this example should require adding cells in every higher dimension. I'd appreciate if you could explain this to me or if you could point me to such an explanation. All the better if it talks through what's happening with the geometry—how exactly does twisty-gluing that first disk into the single loop create a sphere(?) that then needs to get killed? And how does killing this hole lead to homotopy in the next higher dimension, etc.?

Answer 1

I think it's much easier to visualize $B \mathbb{Z}_2$ first (this is just alternate notation for $K(\mathbb{Z}_2, 1)$, emphasizing the classifying space aspect). You'll end up constructing infinite real projective space $\mathbb{RP}^{\infty}$. The first three steps of the construction are the easiest and they go like this:

1. Start with a point. Think of it as two paired points which are identified (no, really), or equivalently as the $0$-sphere $S^0$ modulo the antipodal action of $\mathbb{Z}_2$, or $\mathbb{RP}^0$.
2. Attach a $1$-cell (a loop from the point to itself) to get the generator of $\mathbb{Z}_2$. At this point you have a circle. Think of it as two paired points with two paired edges going between them, which again are identified, or equivalently as the $1$-sphere $S^1$ modulo the antipodal action of $\mathbb{Z}_2$, or $\mathbb{RP}^1$.
3. Attach a $2$-cell to kill twice the generator of $\mathbb{Z}_2$. This is why we doubled everything above: this is much easier to visualize if you just think of it as the top hemisphere of a $2$-sphere, with boundary the $1$-sphere from the previous step. Since we're doubling everything, we'll also add in the bottom hemisphere, so we get the $2$-sphere $S^2$ modulo the antipodal action of $\mathbb{Z}_2$, or $\mathbb{RP}^2$.

At this point we've already constructed a space with the correct fundamental group, so why don't we just stop here? The problem is that $\mathbb{RP}^2$ has the same higher homotopy as its universal cover $S^2$, hence $\pi_2 \mathbb{RP}^2 \cong \mathbb{Z}$. This new higher homotopy is caused by the $2$-cell we inserted to kill twice the generator of $\mathbb{Z}_2$, and so we need to attach a $3$-cell to kill it. The next step is harder to visualize.

4. Attach a $3$-cell to kill the generator of $\pi_2 \mathbb{RP}^2$. You can visualize this as the interior of the $2$-sphere from the previous step, sitting inside $\mathbb{R}^3$. But again, since we're doubling everything, we need to pair this with another $3$-cell. You should think of this as the exterior of the $2$-sphere, sitting inside $\mathbb{R}^3$, together with the point at infinity in $\mathbb{R}^3 \cup \{ \infty \} = S^3$. In other words, we get the $3$-sphere $S^3$ modulo the antipodal action of $\mathbb{Z}_2$, or $\mathbb{RP}^3$.

And so on. At each step we'll end up introducing new higher homotopy in exactly one degree higher than in the previous step, we'll attach exactly one new cell in the next degree to kill it, and we'll get the next sphere modulo the antipodal action of $\mathbb{Z}_2$, which is the next real projective space. At the end of this whole process we'll get the infinite sphere $S^{\infty}$ modulo the antipodal action of $\mathbb{Z}_2$, which is $\mathbb{RP}^{\infty}$.

For $B \mathbb{Z}_n$ for higher $n$ the construction is similar but more complicated. Instead of infinite real projective space you're aiming for an infinite lens space. One necessary complication here is that when $n > 2$, $\mathbb{Z}_n$ can't act freely on an even-dimensional sphere, so if any spheres appear in this construction they must be odd-dimensional.

Answer 2

$K(Z_5,1)$ is infinite Lens space.So it is actually $S^{\infty}/Z_5$.

Another method: Since $\pi_1(K(Z_5,1)) =Z_5=<a|a^5 =1>$.Then we can construct $X$ as follows:

Take one circle.Attach one $2$-cell with attaching map $z \rightarrow z^5$.Now attached $n+1$ -cell to cancel $\pi_n$ for all $n>1$.This $X$ will serve one model for $K(Z_5,1)$.