Why is the second cohomology group of $X=\mathbb{R}P^2$ with $\mathbb{Z}$-coefficients $\mathbb{Z}_2$? We can put the usual $\Delta$-structure on $X$ with two vertices, three $1$-simplices, say $a$, $b$, and $c$, and two $2$-simplices, $U$ and $L$.
The cochain group $C^2=\mathrm{Hom}(\langle U,L\rangle,\mathbb{Z})$ is generated by two maps $\mu$ and $\lambda$, where $\mu(U)=1$, $\mu(L)=0$, and $\lambda(L)=1$, $\lambda(U)=0$. We can define a similar basis $\alpha,\beta,\gamma$ for $C^1=\mathrm{Hom}(\langle a,b,c\rangle,\mathbb{Z})$. The image of $\delta_1$ into $C^2$ is generated by $\delta_1(\alpha)=-\mu+\lambda$, $\delta_1(\beta)=\mu-\lambda$, and $\delta_1(\gamma)=\mu+\lambda$. Then $$ H^2(X,\mathbb{Z})=\ker\delta_2/\mathrm{im}\delta_1=\langle\mu,\lambda\rangle/\langle -\mu+\lambda,\mu-\lambda,\mu+\lambda\rangle=\langle \mu+\lambda,\lambda\rangle/\langle \mu+\lambda,\mu-\lambda\rangle $$ I think this is isomorphic to $\langle\lambda\rangle/\langle\mu-\lambda\rangle$. Why is it $\mathbb{Z}_2$, and not $\mathbb{Z}$?
If you use cw complexes its clearer. Then there is one generator in each dimension and the chain complex is $\mathbb{Z} \rightarrow^2 \mathbb{Z}\rightarrow^0 \mathbb{Z}$ and so the cochain complex is the reverse, and you get $\mathbb{Z}_2$.
Geometrically of you look at it as the top half of a sphere with opposite points of the equator identified then the boundary of the hemsiphere is the equator, which because of the identification is 2 times a 1-cycle.
Yet another perspective is the the surface is not orientable, now $H^2$ is generated by the function dual to a single 2-simplex, but due to non-orientiblilty there is a chain which returns to the same simplex with a different orientation and so this generating function is equivalent with its negative.