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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.

YoYo12
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2 Answers2

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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.

Christoph
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  • $U$ is homeomorphic to $\mathbb{R}^d$ by $(x_1,\dots,x_d)\mapsto [x_1:\dots:x_d:1]$.

  • $V$ does not have homotopy type of a sphere. It is homotopic to a projective space one dimension down (your link says it too). The easiest way is to use the homotopy $[x_0:\dots:x_{d-1},x_d]\mapsto [x_0:\dots:x_{d-1}:tx_d]$.

  • $U\cap V$ has the homotopy type of an $S^{d-1}$ since, in the identification of $U$ with $\mathbb{R}^d$ above, the effect of intersecting with $V$ is to remove the origin.

user10354138
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