De - Rham Cohomology calculation via Mayer - Vietoris Sequences

Question

How can we show that:

$b_1(\mathbb{RP^3})=b_1(\mathbb{RP^2})$ and $b_2(\mathbb{RP^3})=b_2(\mathbb{RP^2})$

I think it is just repeated use of deformation retracts, and Poincare's Lemma to fit in the correct pieces to the Mayer - Vietoris Sequence, but I am still quite weak at computing the Betti numbers for a specific cohomology class. Can anyone show me how one can show the above is equal to each other?

EDIT: $b_k(M)$ is the $k^{\text{th}}$ Betti number of an $n$ - dimensional manifold $M$, and $\mathbb{RP^n}$ is the real projective space of dimension $n$.

My thoughts: If we setup the long exact sequence we have:

$$0\rightarrow H_{dR}^0(\mathbb{RP}^3)\rightarrow H_{dR}^0(U) \oplus H_{dR}^0(V)\rightarrow H_{dR}^0(V\cap U)\rightarrow H_{dR}^1(\mathbb{RP}^3)\rightarrow H_{dR}^1(U) \oplus H_{dR}^1(V)\rightarrow H_{dR}^1(V\cap U)\rightarrow H_{dR}^2(\mathbb{RP}^3)\rightarrow H_{dR}^2(U) \oplus H_{dR}^2(V)\rightarrow H_{dR}^2(V\cap U)\rightarrow $$ $$ H_{dR}^3(\mathbb{RP}^3)\rightarrow H_{dR}^3(U) \oplus H_{dR}^3(V)\rightarrow ...$$

Now assuming the sequence above is exact, what I am having trouble seeing is, the exact values of the dimensions for the Cohomology classes of open sets $U$ and $V,$ where I defined the open sets $U$ as the real projective space of dimension three excluding some point $x,$ and $V$ is a $3 $ - dimensional sphere containing the point $x$. It really seems confusing finding the $H_{dR}^k(U\cup V)$ and $H_{dR}^k(U \cap V)$, for $k = 1, 2 ,3.$

Answer 1

Real Projective space is not some abstract space, and you can get everything you need to know about the cohomology of the space from e.g. https://topospaces.subwiki.org/wiki/Cohomology_of_real_projective_space.

The calculation itself is almost everywhere, see e.g. Homology group of Real projective plane where the first answer is a nice demo of how to calculate the homology using Mayer-Vietoris sequence. After knowing the homology of the space, we can get the cohomology from the universal coefficient theorem. (Or we can just modify the calculation to calculate the cohomology group directly.) See also e.g. Hatcher's book for the calculation using cellular (co)homology.

Answer 2

$\mathbb{R} P^3$ is the quotient of $S^3$ by the antipodal map $x\mapsto -x$. Choose a point $x\in S^3$ and consider $S^2\subset S^3$ as the equator with respect to the poles $x$ and $-x$. Consider the two open subsets $U$ and $V$ of $S^3$, where $U$ is the open lower hemisphere of $S^3$ and $S^2\subset V$ is a small $\epsilon$-neighborhood of the equator. Since the quotient map $q:S^3\to \mathbb{R} P^3$ is a quotient by a group action (i.e. $\mathbb{Z}/2$), it is an open mapping (in fact $q$ is a covering projection). Hence, $q(U)$, $q(V)$ give an open cover of $\mathbb{R} P^3$. Moreover, after a bit of thought, you see that $q(U)$ deformation retracts to a point, $q(V)\simeq \mathbb{R} P^2\subset \mathbb{R} P^3$ and $q(U)\cap q(V)\simeq S^2$ (the equator).

The Mayer-Vietoris sequence with this open cover gives you the sequence $$... \to H_{dR}^{*-1}(S^2)\to H_{dR}^*(\mathbb{R} P^3)\to H_{dR}^*(\mathbb{R} P^2)\oplus H^*_{dR}({\rm pt})\to H_{dR}^{*}(S^2) \to ... $$ Since $H^*_{dR}({\rm pt})\cong 0$ for $\ast>0$, and $H^0_{dR}(S^2)\cong H^0(\mathbb{R} P^3)\cong H^0(\mathbb{R} P^2)\cong \mathbb{R}$, you get the two sequences $$\mathbb{R} \to \mathbb{R}\to H^1_{dR}(\mathbb{R} P^3)\to H^1_{dR}(\mathbb{R} P^2) \to 0$$ and $$0\to H^2_{dR}(\mathbb{R} P^3)\to H^2_{dR}(\mathbb{R} P^2) \to \mathbb{R}\to \mathbb{R}\cong H_{dR}^3(\mathbb{R} P^3)\;.$$ The isomorphism $\mathbb{R}\cong H_{dR}^3(\mathbb{R} P^3)$ uses the fact that odd projective spaces are orientable and hence have fundamental classes. If you can convince yourself that the first map in the first sequence and last map in the last equation are iso's (which is not too hard to do, just thinking about what these maps do) then your done.

Edit:

The connecting homomorphism in the Mayer-Vietoris sequence for de Rham cohomology has a nice description a la Bott and Tu. You can use that to show the two maps in those sequences are iso's, but I realized its a bit easier to just pull the sequence out a bit further. In particular, for the second sequence, you can extend to the right by $$\mathbb{R} \to \mathbb{R}\cong H_{dR}^3(\mathbb{R} P^3) \to H^3_{dR}(\mathbb{R} P^2)\;.$$ But for dimension reasons $H^3(\mathbb{R} P^2)=0$ and, by exactness, that first map is a surjective linear map, hence injective, hence an iso.