Q:

Homotopy type of a disk and homotopy type of a sphere.

According to the following link : Calculation of de Rham complex for real projective space , $ \mathbb{P}^d $ is devided in two open sets : $ U = \{ \ [ x^0 : \dots : x^d ] \ | \ x^d \neq 0 \ \} $ and $ V = \mathbb{P}^d \backslash \{ \ [ 0 : 0 : \dots : 0 : 1 ] \ \} $ such that : $ \mathbb{P}^d = U \cup V $

How to prove that $ U $ has the homotopy type of a disk $ D^d $ ?
How to prove that $ V $ has the homotopy type of a sphere $ S^{d-1} $ ?

Thanks in advance for your help.

A:

There is a homeomorphism $\mathbb R^d \to U$ given by
$$(x^0,\ldots,x^{d-1}) \longmapsto [x^0 : \ldots : x^{d-1} : 1].$$
Hence, $U$ is homeomorphic to $\mathbb R^d$ which is in turn homeomorphic to the disk $D^d$. In particular they all have the same homotopy type, which is just the homotopy type of a point, since these spaces are contractible.
The space $V$ does not have the homotopy type of $S^{d-1}$. It has the homotopy type of $\mathbb P^{d-1}$. To see this, embed $\mathbb P^{d-1}$ into $\mathbb P^d$ by
$$[x^0 : \ldots : x^{d-1}] \longmapsto [x^0 : \ldots : x^{d-1} : 0].$$
You can now deformation retract $V$ onto $\mathbb P^{d-1}$ using the homotopy $V\times I\to V$ given by
$$
([x^0 : \ldots : x^{d-1} : x^d], t) \longmapsto \left[x^0 : \ldots : x^{d-1} : (1-t) x^d\right]
$$
What does have the homotopy type of $S^{d-1}$ is the intersection $U\cap V$, which is homeomophic to $\mathbb R^d \setminus \{0\}$ under the homeomorphism I gave above.

