1

I think the question is equal to whether a n-manifold has a n-submanifold which is compact in n-manifold. I feel there is not such manifold, but I don't know how to prove it. In fact, I just need some reference about the proof. Thank you.

Farmer
  • 1,535
  • 3
    What about the disjoint union of two $n$-spheres? – Oscar Randal-Williams Feb 03 '14 at 03:48
  • You might consider reading Hatcher's book: Algebraic Topology available online: http://www.math.cornell.edu/~hatcher/AT/ATpage.html. Chapter 2 should have what you are looking for. – Neil Hoffman Feb 03 '14 at 03:50
  • In response to your first sentence, you may want to consider that any open subset of an n-manifold is an embedded n-submanifold. – Brian Klatt Feb 03 '14 at 05:29
  • @BrianKlatt sorry,i am really careless.If assume the manifold is connected and compact and the submanifold must be compact ,how to proof there is not the n-submanifold?Besides,how to compute the open ball's homolopy group,i don't know how to compute the manifold's homolopy group which's triangulation is infinite or is not compact, – Farmer Feb 03 '14 at 05:51
  • @lanse2pty, any closed $n$-submanifold is clopen. – HJRW Feb 03 '14 at 05:53
  • Please, if you need to correct anything in the body of the question (and if you are «really being careless», you do need to!) then please do. It is no a good idea to leave corrections hid in comments. – Mariano Suárez-Álvarez Feb 03 '14 at 06:00

1 Answers1

4

If you assume the manifold to be connected, the answer is not, there aren't such manifolds. For any closed $n$-manifold $M$, we have $H_n(M; \Bbb Z) = \Bbb Z$ if $M$ is orientable, or $H_n(M; \Bbb Z) = 0$ if it is not orientable. If $M$ is not closed then $H_n(M; \Bbb Z) = 0$. This is definitely a special case of Poincar\'e duality, although it follows from the existence of a handlebody decomposition with only one $n$-handle for closed $M$, or without $n$-handles otherwise. You can look at almost any book about homology theory (e.g. Vick), or about differential topology. Of course, in the disconnected case you can easily construct manifolds which satisfy your condition (take the disjoint union of any two closed, orientable, connected $n$-manifolds), see other comments.

Daniele Zuddas
  • 166
  • 1
    The question does not assume that manifold is connected. If one allows for the disconnected case, then the answer can be yes as Oscar Randal-Williams points out. – Neil Hoffman Feb 03 '14 at 03:56
  • Ops you're right, thank you! I edited the answer to include the disconnected case. – Daniele Zuddas Feb 03 '14 at 05:05
  • sorry,i forget to say it is connected.Are you sure that $H_n(M;Z)=Z_2$if it is not orientable? I think the connected n-manifold's $H_n(M;Z)$is Z or $0$,how does it is $Z_2$?Could you give me a example? – Farmer Feb 03 '14 at 05:38
  • You right sorry! I corrected the mistake in the non-orientable case. – Daniele Zuddas Feb 03 '14 at 05:54