In the follwing we shall use $\mathbb{Z}_2$ coefficients. Then the homology as well as the cohomology groups of $X=\mathbb{R}P^2$ agree and are isomorphic to $\mathbb{Z}_2$ in degree 0,1,2 and zero else. Consider the following triangulation of $X$:
As $a$ is generating the first homology group we may dualize it to an $\alpha$ which is the characteristic function of $a$; that means 1 on $a$ and zero else. My goal is to show that $$\alpha\smile\alpha\neq0$$which then shows that this is the generator of $H^2(X).$ This is true according to Hatchers Algebraic Topology Book on p.208 Example 3.8
To achieve this I showed that this $\alpha$ is the cohomology equivalence class of a 1-cochain $\tilde\alpha$ which is again the characteristic function of $a$. My idea is to show that $$\tilde\alpha\smile\tilde\alpha(Y+Z)\neq0.$$I proved that $Y+Z$ is indeed a 2-chain: by coefficients and identification of $a$. My computation yields \begin{align*}\tilde\alpha\smile \tilde\alpha(Y+Z)&=\tilde\alpha\smile\tilde\alpha(Y)+\tilde\alpha\smile\tilde\alpha(Z)\\&=(\tilde\alpha(Y\circ {}^1\pi)\cdot\tilde\alpha(Y\circ\pi^1))+(\tilde\alpha(Z\circ{}^1\pi)\cdot\tilde\alpha(Z\circ\pi^1))\\&=\tilde\alpha(v)\cdot\tilde\alpha(a)+\tilde\alpha(w)\cdot\tilde\alpha(a)=0\cdot 1+0\cdot 1=0.\end{align*}where ${}^1\pi$ projects onto the first 2 entries of a given $t=(t_0,t_1,t_2)\in\mathbb{R}^3$ and $\pi^1$ projects onto the last 2 entries of a given $t$. This has to be wrong because we see by clearly that $\tilde\alpha\smile\tilde\alpha$ is in fact zero. My question is: Is it possible to fix this issue? - Did I do some mistake? If not, is there a way around this issue to prove my goal? I did a similar example on the torus which worked out well and I dont know where my mistake is for $X$. I am grateful for any comments or recommendations since I really don't know what else I should do to prove this, thanks in advance.
Your $\tilde\alpha$ is not actually a cocycle. To get a cocycle you need something that vanishes on boundaries, so if it sends $a$ to $1$, it also needs to send $v+w$ to $1$, so that it vanishes on $\partial Y$ and $\partial Z$. So exactly one of $\tilde\alpha(v)$ and $\tilde\alpha(w)$ should be $1$, and then $\tilde\alpha\smile \tilde\alpha(Y+Z)$ will also be $1$.