2
$\begingroup$

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.

Improve this question
$\endgroup$
2
  • 4
    $\begingroup$ You would first want to show that any element $\alpha$ of $H_2(X;\mathbb{Z})$ can be realized by some map of a closed, oriented surface into $X$. If you're using simplicial homology then $\alpha$ is a bunch of triangles which glue together in $X$ and has no boundary. You need to resolve this object at the vertices where it may not be like a surface. Now use the classification of surfaces and kill off the generators ($\pi_1(X)=0$) of the fundamental group of this surface. Homologically, this means that $\alpha$ is the sum of spheres mapping into $X$ and then use $\pi_2(X)=0$. $\endgroup$
    – Somnath Basu
    Commented Feb 1, 2011 at 23:59
  • $\begingroup$ It is not true that there is a simpler proof, because the question basically is the Hurewicz theorem. That is, the full version of the Hurewicz theorem looks stronger, but this special case already requires basically all of the ideas. $\endgroup$
    – Greg Kuperberg
    Commented Feb 2, 2011 at 21:03

2 Answers 2

Reset to default
11
$\begingroup$

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.)

Improve this answer
$\endgroup$
1
  • $\begingroup$ Yes, this is certainly a very elementary way of explaining this fact. Thanks! $\endgroup$
    – Isaac Goldbring
    Commented Feb 2, 2011 at 16:39
1
$\begingroup$

You can build a weak homotopy equivalence $K \to X$ from a CW complex with $2$-skeleton $K_2 = *$. Then cellular homology gives the result, provided your homology respects weak equivalence.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .