Let $p \in \text {int} (D^{2n}).$ Then show that $\mathbb C P^n - p \simeq \mathbb C P^{n-1}.$
I know that $\mathbb C P^n \approx \mathbb CP^{n-1} \sqcup_q D^{2n},$ where $q : S^{2n-1} \longrightarrow \mathbb C P^{n-1} \approx S^{2n-1}/S^1$ that identifies points on $S^{2n-1}$ which are $S^1$ multiples of each other. So if we take $p \in \text {int} (D^{2n})$ then clearly $\mathbb C P^{n-1} \subseteq \mathbb C P^n - p$ and $D^{2n} - p \simeq S^{2n-1}.$ From here can it be concluded that $\mathbb C P^n - p \simeq \mathbb C P^{n-1}\ $?
Any help in this regard would be much appreciated. Thanks for your time.
Manifolds are homogenous, meaning that there is a homeomorphism taking any one point to any other. Therefore, we might as well assume $p$ is the center of $D^{2n}$. If we remove a point from the center of a disk, we may radially scale to deformation retract onto the boundary sphere. We may extend this by the identity everywhere to conclude $\mathbb{C}P^n -p$ deformation retracts onto $\mathbb{C}P^{n-1}$.