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I computed some homology groups, could you tell me if I got it right? Thanks.

(i) Let $M$ be the Möbius band. Then $H_0(M, \mathbf{Z}_2) = H_1(M, \mathbf{Z}_2) = \mathbf{Z}_2$ and $H_n(M, \mathbf{Z}_2) = 0$ otherwise.

(ii) $H_0 ( S^1, \mathbf{Z}_2) = H_1(S^1, \mathbf{Z}_2) = \mathbf{Z}_2$ and $H_n(S^1, \mathbf{Z}_2) = 0$ otherwise.

So we cannot distinguish between the Möbius band and $S^1$ in $\mathbf{Z}_2$, even though they are not homeomorphic.

(iii) $H_0 (S^2, \mathbf{Z}_2) = \mathbf{Z}_2 $, $H_2 ( S^2, \mathbf{Z}_2) = \mathbf{Z}_2$ and $H_n(S^2, \mathbf{Z}_2) = 0$ otherwise.

(iv) Let $T = S^1 \times S^1$. Then $H_0(T, \mathbf{Z}_2) = \mathbf{Z}_2$, $H_1(T, \mathbf{Z}_2) = \mathbf{Z}_2 \oplus \mathbf{Z}_2$ and $H_2(T, \mathbf{Z}_2) = \mathbf{Z}_2$. $H_n(T,\mathbf{Z}_2) = 0$ otherwise.

So we see that we can still distinguish some non-homeomorphic spaces.

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1 Answer 1

up vote 3 down vote accepted

(i),(ii) are correct. You should prove (or believe) that the Möbius band and the circle are homotopic, so whatever homotopy invariant you are looking at, they should be the same!

(iii) Again correct. Try and convince yourself what $H_i(S^n;\mathbb{Z}/p\mathbb{Z})$ should be now

(iv) Correct again.

I'm not sure there is much more to add!

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Don't worry about adding anything : ) I'm glad to know that everything is correct. –  Matt N. Aug 29 '12 at 8:35
4  
"the Moebius band and the circle are homotopic, so whatever topological invariant you are looking at, they should be the same" I think you should replace "topological invariant" with "homotopy invariant", as the two spaces are not homeomorphic. –  Alex Becker Aug 29 '12 at 8:36
    
(it won't let me accept, so I'll have to accept later) –  Matt N. Aug 29 '12 at 8:36
    
@Alex - Dear Alex, you are correct and I have edited it! –  Juan S Aug 29 '12 at 8:39

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