Can one prove that $\Bbb{CP}^\infty$ is a $K(\Bbb Z, 2)$ without invoking the long exact sequence of a fibration?

Question

In trying to remind myself why $\Bbb{CP}^\infty$ is a $K(\Bbb Z, 2)$, the natural argument that comes to mind is to take the long exact sequence associated to the fibration $S^1 \rightarrow S^\infty \rightarrow \Bbb{CP}^\infty$. Since $S^\infty$ is contractible, the desired result follows. But it feels to me like there should be an "easier" way of doing this; perhaps just from knowing the cell decomposition of $\Bbb{CP}^\infty$. Is there?

I'm aware that this is a bit of a silly question, and will be completely satisfied with "No, that's pretty much the easiest/most elementary way to do it."

Answer 1

"If the long exact sequence of a fibration did not exist it would have been necessary to invent it."

Supposing you know the fibration $S^1 \to S^\infty \to \Bbb{CP}^\infty$ but are unaware of the long exact sequence of a fibration. Here's how you might try to prove the desired result.

Well, I know $\pi_0 = \pi_1 = 0$ by cellular approximation, and $\pi_2 = \Bbb Z$ by Hurewicz. So, ok, suppose I have a map $S^3 \to \Bbb{CP}^\infty$. Well, I want to lift it to $S^\infty$ - I know that's contractible. When am I allowed to lift it? Well, let's cheat a bit. Let's consider the map from $S^3$ as a map from $D^3$ that scrunches the boundary to a point. That's contractible, so I can lift that to $S^\infty$. I know that the map, restricted to $\partial D^3$, lives in the fiber $S^1$... But $\pi_2(S^1) = 0$, so I can probably null-homotope the boundary to get an actual lift of $S^3$... So that the map is null-homotopic.

And thus, when we considered the map from the boundary of the ball, we have invented the boundary map $\pi_3(\Bbb{CP}^\infty) \to \pi_2(S^1)$, just from doing our best to use the tools at our disposal.

(This might seem a little ad hoc, but there have been situations where I've done essentially this same idea because it's all I could think of doing, where my domain was something other than $S^n$ and I had a map to something fitting into a fibration.)