I don't understand a passage in my lecture notes. Here is the passage with my questions added in italics:
$H^\ast(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$ is $\mathbb{Z}/2\mathbb{Z}$ in degrees $1$ and $2$. Let $\gamma$ be the a generator of $H^1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$. Let us compute $\gamma \cup \gamma$. Recall the presentation $\mathbb{R}P^2$ as a circle (It should be "disk" and not "circle“, right?) with the two halves of the boundary identified. We fix a base point $\ast$. Let $\ast_1$ be the constant $1$-simplex and $\ast_2$ the constant $2$-simplex at the base point $\ast$. Let $c$ be the $1$-simplex that is half the boundary of the disk, it is easy to see this is a generator for $\pi_1(\mathbb{R}P^2, \ast)$ and thus for $H_1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$. (How do I see that $c$ is a generator for $\pi_1(\mathbb{R}P^2, \ast)$?). Lastly, let $s$ be the $2$-simplex that maps onto the disk homeomorphically, with boundary $2\gamma -\ast_1$. (This should read "$2c-\ast_1$" and not "$2\gamma -\ast_1$"?). It follows that $s-\ast_2$ is a generator of $H_2(\mathbb{R}P^2)$. (Why does that follow?).
We then must have $\gamma(c)=1$ and $\gamma(\ast_1)=0$. ($\gamma(\ast_1)=0$ follows from $\ast_1=\partial \ast_2$ and $\gamma(c)=1$ from $\gamma\neq 0$ and $c$ being a generator of $H_1(\mathbb{R}P^2;\mathbb{Z}/2\mathbb{Z})$, correct?). We therefore find $\gamma \cup \gamma (s)=1$ and $\gamma \cup \gamma (\ast_2)=0$. We still need to show that $\gamma\cup \gamma$ is not a coboundary (Why is this not obvious?).