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For the projective space the cell decomposition is $e_0 \cup \dots \cup e_n$ and the attaching map is $a_1 a_1 \dots a_k a_k$ for the $k$-th cell. So for $k \leq n$ I thought that this means that the boundary map between the cellular chain groups is multiplication by two because the boundary map of the $k$-th cell coincides with the attaching map. So I thought $im \partial_k = \langle a_1 a_1 + \dots + a_k a_k \rangle = 2 \langle a_1 + \dots + a_k \rangle \cong 2 \mathbb{Z}$.

Why is this wrong? In the example on p. 144 in Hatcher the chain complex looks like this

$$ \dots \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z}\xrightarrow{2} \mathbb{Z} \dots$$

How can the boundary map be the $0$ map? Many thanks for your help.

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Consider the case of $\mathbb RP^1=S^1$... –  Grigory M Aug 25 '11 at 9:12
    
Yes I did that already, it's multiplication by $2$! –  Rudy the Reindeer Aug 25 '11 at 9:32
3  
No, it is not, you're missing signs ($H_0$ can never have torsion, btw). –  Grigory M Aug 25 '11 at 9:37
    
Oooh! So half of it is glued around one direction and the second half of it is glued around the opposite direction so that it becomes the $0$ map? –  Rudy the Reindeer Aug 25 '11 at 9:49

1 Answer 1

you can use of local degree and the fact degree of reflection is $(-1)^{n+1}$ .or if you use $I^n$, $\partial_k=<a_1-(-1)^{k+1}a_1+...>$ because degree of reflection is $(-1)^{n+1}$

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