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In they use the Universtal coefficient theorem

$$0 \rightarrow \mbox{Ext}(H_{i-1}(X, \mathbb{Z}),A)\rightarrow H^i(X,A)\rightarrow\mbox{Hom}(H_i(X, \mathbb{Z}),A)\rightarrow 0$$

to determine the cohomology of $RP^n$ with coefficient in the field $\mathbb Z_2$ which gives that: $$\forall i = 0 \ldots n , \ H^i (RP^n; \mathbb Z_2) = \mathbb Z_2$$

But in Hatcher page 198, he says that with field coefficients cohomology is the dual of homology that is $H^k(X,\mathbb Z_2)=Hom(H_k(X,\mathbb Z_2);\mathbb Z_2)$. I applied this to get

$$H^i(RP^n; \mathbb Z_2) = \begin{cases} \mathbb Z_2 & i = 0 \mbox{ and}, 0<i\leq n, \ i\ \mbox{odd,} \\ 0 & \mbox{else.} \end{cases}$$

I thought it was a problem in the use of the Universal coefficients theorem taking $\mathbb Z_2$ as a group not as a field but in Hatcher page 199 he says that when the field is $\mathbb Z_p$ or $\mathbb Q$ it does not matter! I'm confused.. thanks for help

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(Re: second part) how do you compute $H_i(RP^n;\mathbb Z/2)$? – Grigory M Nov 27 '11 at 19:55
$H_i(\mathbb{RP}^n;\mathbb Z_2)=\mathbb Z_2$ for all $0\leq i\leq n$. – Grumpy Parsnip Nov 27 '11 at 19:59
@palio: You need to use the universal coefficient theorem. You can't just tensor with $\mathbb Z_2$. – Grumpy Parsnip Nov 27 '11 at 20:00
@palio That's incorrect. (Have you read the Wikipedia page you're linking to?..) – Grigory M Nov 27 '11 at 20:00
@Grigory M : ok i see that i did not take the Tor functor into account.. now it is clear thankyou very much!!! – palio Nov 27 '11 at 20:08

(Just not to leave this unanswered.)

Wikipedia is right, $H^i(\mathbb RP^n;\mathbb Z/2)$ is $\mathbb Z/2$ for $0\le i\le n$. This agrees with Hatcher, because $H_i(\mathbb RP^n;\mathbb Z/2)$ is also $\mathbb Z/2$ for $0\le i\le n$.

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