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Let $K$ denote the $2$ dimentional projective plane $RP^2$ . Then we have $$H_0(K) \cong Z$$ $$H_1(K) \cong Z_2$$ $$H_2(K)=0$$
Let $b_i$ denote the Betti number of $K$ , $b_i=rank(H_i(K))$, then we have $$E=b_0-b_1+b_2$$
We know that the Euler number $E$ of $K$ is $1$ , $b_0=1$, $b_2=0$ . So we must have $b_1=0$ , which means that $rank(Z_2)=0\neq1$ . However , $rank(Z_2)$ should be $1$ since $<1>=Z_2$ .

There must be something wong in the argument above , but I can't see where it is . Any help would be very appreciate .

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The rank of a module $M$ is the rank of its free part, or equivalently the maximal number of linearly independent elements in $M$. $\mathbb{Z}_2$ has no linearly independent elements, since $2 \cdot 1 = 0$ in $\mathbb{Z}_2$, and hence its rank is 0.

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    $\begingroup$ Alternatively the rank of an abelian group ($\mathbb Z -$module) $A$ is the largest number $n$ such that there is an injective map $\mathbb Z^n \rightarrow A$ $\endgroup$ Feb 9 '20 at 13:27
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You are working in homology with coefficients in $\mathbb Z$, where the chain groups and homology groups are modules over $\mathbb Z$. The rank of a $\mathbb Z$-module is equal to the maximum rank of a free submodule over $\mathbb Z$. So no, $\text{rank}(\mathbb Z_2)=0$, not $1$, because the only free submodule over $\mathbb Z$ of $\mathbb Z_2$ is trivial.

Perhaps your are also trying to think about homology with $\mathbb Z_2$ coefficients? If so, you get a different calculation with a consistent outcome: $$H_0(K;\mathbb Z_2) \cong \mathbb Z_2$$ $$H_1(K;\mathbb Z_2) \cong \mathbb Z_2$$ $$H_2(K;\mathbb Z_2) \cong \mathbb Z_2$$ all of which have rank $1$ as modules over $\mathbb Z_2$, and so the Euler characteristic over $\mathbb Z_2$ is $1-1+1=1 \in \mathbb Z_2$, which of course is equal to the reduction modulo $2$ of $1 \in \mathbb Z$.

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