I am looking at the real projective plane and I am supposed to show that is possesses the structure $\mathbb{R}P^n = e_0 \cup\cdots\cup e_n$.
Well, I know that $\mathbb{R}P^n = S^n/(x \sim -x)$
I also know that the canonical quotient map $\phi: \partial D^n = S^{n-1} \rightarrow S^{n-1}/(x \sim \ -x)$ is the adjunction map.
Now, I need to show that $S^n/(x \sim \ -x) = (S^{n-1} \cup D^n)/(x \sim \ \phi(x))$. Now I thought that we can simplify the right-hand side, as the equivalence relation is independent from $\operatorname{int}(D^n)$, such that $(S^{n-1} \cup D^n)/(x \sim \ \phi(x)) = (S^{n-1} )/(x \sim \ \phi(x)) \cup \operatorname{int} D^{n}$. Is this reasoning correct? Cause this would help me a lot, as now I know that this is the same as $\mathbb{R}P^{n-1} \cup \operatorname{int} D^{n} =\mathbb{R}P^{n-1} \cup e_n $, right?
My problem is that I just cannot write down the homeomorphism between $S^n/(x \sim \ -x) = (S^{n-1} \cup D^n)/(x \sim \ \phi(x))$. In my opinion these two spaces are not homeomorphic, cause if I have a 1d subspace let's say in $\mathbb{R}^3$(which is a line), then I can project this one into the plane just by tilting. So the only thing I need to remember is: How much did I tilt and therefore I don't see why you need $\operatorname{int}(D^n)$ if you are only interested in : How much did you tilt?
Starting from $S^n / (x \sim (-x))$, remove the open lower hemisphere, leaving just the closed upper hemisphere which I'll denote $S^n_+$, whose boundary is $S^{n-1}$. So you get a homeomorphism $$S^n / \{x \sim (-x)\} \approx S^n_+ / \{x \sim (-x) \,| \, x \in S^{n-1}\} $$ Now just replace the closed upper hemisphere $S^n_+$ with the closed disc $D^n$ to which it is homeomorphic.