I am trying to understand the computation of a De Rham Cohomology of a Real Projective Space. When we cover $ℙ^d$ with the sets $U=\{[x^0,\dots,x^d]: x^d≠0\}$ and $V=\mathbb{P}^d\smallsetminus \{[0,...,0,1]\}$, i notice that we use the fact that $U$ has the homotopy type of a disk (so it is contractible to a point), $V$ has the homotopy type of $ℙ^{d−1}$ and $U∩V$ has the homotopy type of $S^{d−1}$. Could someone give me a very clear explanation as to why this is true?
I am not very familiar with Homotopy Theory and hence would need a very clear explanation.
Thanks.
Well $(t, [x_0 :...:x_d]) \mapsto [tx_0 :tx_1 :...: x_d]$ is a homotopy between $id_U$ and the constant map $U\to U$ with value $[0:...:0:1]$. That it is continuous is clear by the universal property of the quotient; this explains why $U$ is contractible.
For $V$ you have a projection $V\to \mathbb{P}^{d-1}$ defined by $[x_0 :...: x_d] \mapsto [x_0 :...: x_{d-1}]$. It's well defined because you can't have all $x_0,...,x_{d-1}$ $0$ at the same time (by definition of $V$); and clearly continuous. Then you have a "converse" $\mathbb{P}^{d-1} \to V$ defined by $[x_0:...:x_{d-1}]\mapsto [x_0:...:x_{d-1}: 0]$. You have to show that these maps form a homotopy equivalence; it should not be too complicated so I'll leave this computation to you.
Finally, $U\cap V$. Being in this set means your $d$ coordinate is nonzero, but you have another nonzero coordinate. A way to see that this is homotopy equivalent to $S^{d-1}$ is to set $f: S^{d-1} \to U\cap V$, $x\mapsto [x_0 :....: x_{d-1} : 1]$ and $g : U\cap V \to S^{d-1}$ defined by $[x_0:...:x_{d-1} : 1]\mapsto \frac{x}{||x||}$. Again, you have to show that these maps form a homotopy equivalence but this shouldn't be too complicated either and I'll leave those details to you as well.