I am looking for an "easy" proof of the following statement: Suppose that $X$ is a simply connected space for which $\pi\_2(X)=0$. Then $H\_2(X)=0$ as well.


I know that one can use the Hurewicz theorem to prove this, but I feel like there must be a simpler proof.



vanishing of $\pi_2$ and $H_2$
Claim 1: If $x\in H\_2X$ is a homology class, then there exists a 2-dimensional CW complex $K$ and a map $f:K\to X$, such that $x$ is in the image of $f\_\*:H\_2K\to H\_2X$.


Claim 2: If $X$ is simply connected with $\pi\_2X=0$, then every map $K\to X$ from a 2-dimensional CW complex is null homotopic.


Claim 1 can be proved by taking $K$ to be a finite union of triangles attached together along suitable edges, built by considering an explicit cocycle representing $x$. Claim 2 is easy.


Of course, this is just the $n=3$ case of the standard proof of "easy Hurewicz" (that $\pi\_kX=0$ for $k< n$ implies $H\_kX=0$ for $k<n$; non-easy Hurewicz is the statement relating the non-trivial groups in dimension $n$, but you don't want that part.)


