Considering the CW complex structure on $\mathbb{R}P^{2}$ consisting of one $0$-cell, one $1$-cell and one $2$-cell. Then the cellular chain complex of $\mathbb{R}P^{2}$ in $\mathbb{Z}/2$-coefficient is given by $$\mathbb{Z}/2\cdot e^{2} \longrightarrow \mathbb{Z}/2\cdot e^{1} \longrightarrow \mathbb{Z}/2\cdot e^{0}$$ where $e^{i}$ denotes the $i$-cell and all differentials are zero.
Applying the functor $\mathrm{Hom}(-,\,\mathbb{Z}/2)$, we can get the cellular cochain complex of $\mathbb{R}P^{2}$ in $\mathbb{Z}/2$-coefficient: $$\mathbb{Z}/2\cdot \epsilon ^{2} \longleftarrow \mathbb{Z}/2\cdot \epsilon ^{1} \longleftarrow \mathbb{Z}/2\cdot \epsilon ^{0}$$ where $\epsilon^{i}$ can be thought as the function sending $e^{i}$ to $1$.
Since differentials are all zero, the cellular cohomology of $\mathbb{R}P^{2}$ in $\mathbb{Z}/2$-coefficient is also generated by $\epsilon^{i}$'s.
My question is what is the cup product $\epsilon ^{1}\cup \epsilon ^{1}\in H^{2}(\mathbb{R}P^{2};\,\mathbb{Z}/2)$? It could be $0$ or $\epsilon ^{2}$.
The cup product can be viewed as the composition of $$H^{1}(\mathbb{R}P^{2};\,\mathbb{Z}/2)\otimes H^{1}(\mathbb{R}P^{2};\,\mathbb{Z}/2) \longrightarrow H^{2}(\mathbb{R}P^{2}\times \mathbb{R}P^{2};\,\mathbb{Z}/2)\longrightarrow H^{2}(\mathbb{R}P^{2};\,\mathbb{Z}/2)$$
The first map sends $\epsilon^{1}\otimes\epsilon^{1}$ to the 2-cochain on $\mathbb{R}P^{2}\times \mathbb{R}P^{2}$ that takes value $1$ on $e^{1}\times e^{1}$ and $0$ on $e^{2}\times e^{0}$ and $e^{0}\times e^{2}$.
The second map is dual to the cellular approximation of the diagonal map $$\Delta:\,\mathbb{R}P^{2}\longrightarrow\mathbb{R}P^{2}\times \mathbb{R}P^{2}$$ However, I don't understand this map. If it sends $e^{2}$ to $e^{1}\times e^{1}$, then we have $\epsilon ^{1}\cup \epsilon ^{1}=\epsilon^{2}$, otherwise we will have $\epsilon ^{1}\cup \epsilon ^{1}=0$.
I really appreciate if someone can give me some explanations on this cup product via cellular cochains or some geometry behind this product.