2

So there is something that I don't quite understand in Hatcher's proof.

enter image description here

I would appreciate it if something could tell me why $H_n(M:\mathbb{Z}_p)$ would be larger than the $\mathbb{Z}_p$ coming from $H_n(M;\mathbb{Z})$. Thanks.

Enigma
  • 3,909
  • 3
    possibly duplicate of http://math.stackexchange.com/questions/1532558/m-orientable-implies-h-n-1m-mathbbz-is-free-abelian-group/1532791#1532791 – Anubhav Mukherjee Feb 15 '16 at 17:09
  • @Anubhav.K: This is not a duplicate, since the question that you pointed to only treats the orientable case, which is half of the problem that we have here. May I suggest that you retract the close vote? – Alex M. Feb 15 '16 at 21:40

2 Answers2

2

By the universal coefficient theorem $$ H_n(M; \def\Z{\mathbf Z}\Z_p) \cong \Z_p \otimes H_n(M;\mathbf Z) \oplus \mathrm{Tor}(H_{n-1}(M;\mathbf Z), \Z_p) $$ If $H_{n-1}(M;\mathbf Z)$ had $p$-torsion, then $\mathrm{Tor}(H_{n-1}(M;\mathbf Z), \Z_p) \ne 0$, hence $H_n(M; \Z_p)$ would be larger then $\Z_p \cong \Z_p \otimes H_n(M;\Z)$, which is not possible.

martini
  • 84,101
0

For an alternative proof in the orientable case consider Poincaré duality:

$$ H_{n-1}M\cong H^1M \cong Hom(H_1M,\mathbb Z) \cong H_1M /tors $$

Daniel Valenzuela
  • 6,305
  • 12
  • 20