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In http://en.wikipedia.org/wiki/Universal_coefficient_theorem 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 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 $H_i(RP^n;\mathbb Z/2)$?2011-11-27
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    @Grigory M : By tensoring $H_i(RP^n;\mathbb Z)$ by $\mathbb Z_2$.2011-11-27
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    $H_i(\mathbb{RP}^n;\mathbb Z_2)=\mathbb Z_2$ for all $0\leq i\leq n$.2011-11-27
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    @palio: You need to use the universal coefficient theorem. You can't just tensor with $\mathbb Z_2$.2011-11-27
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    @palio That's incorrect. (Have you read the Wikipedia page you're linking to?..)2011-11-27
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    @Grigory M : ok i see that i did not take the Tor functor into account.. now it is clear thankyou very much!!!2011-11-27

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