I'm trying to compute the De Rham cohomology of the manifold
$$M_d=\mathbb{R}^2\setminus\{ p_1\ldots p_d\},$$
where the points $p_1\ldots p_d$ are all distinguished.
This must be a standard exercise in the use of the Mayer-Vietoris sequence, but I run into a difficulty. Namely, taking a suitable open covering, such as
$$M_d=U\cup V,$$
where (up to homeomorphism) $U=M_{d-1}$ and $V=M_1$, and then observing that $M_1$ is homotopically equivalent to the $1$-manifold $\mathbb{S}^1$, I can obtain from the Mayer-Vietoris sequence the following piece of information:
$$h^1(M_d)-h^2(M_d)=h^1(M_{d-1})-h^2(M_{d-1})+1,$$
(where $h^i(\ldots)=\dim H^i(\ldots)$).
What is causing trouble is the presence of those $h^2$-terms. I guess that they should vanish, since $h^2(M_1)$ does because of the said homotopical equivalence with $\mathbb{S}^1$. But,
is it true that the plane with $m$ holes is homotopically equivalent to a manifold of dimension $1$?
Intuitively I would say that this is true: $M_2$ is homotopically equivalent to an $8$, $M_3$ to a curve that does three loops and so on. The problem is that those are not manifolds: the figure $8$, for example, has a singularity in its centre. This leaves me puzzled.
Thank you for reading.
Best Answer
Lets pursue this induction: Suppose that in what follows, at each step, $M_{d-1}\cap M_1$ is $\mathbb{R}^2$ less $d$ distinct points. You have a short exact sequence:
$$ 0 \rightarrow \Omega^*(M_1\cup M_{d-1}) \rightarrow \Omega^*(M_1)\oplus\Omega^*(M_{d-1}) \rightarrow \Omega^*(M_d) \rightarrow 0$$
When $d = 2$, we have the induced long exact sequence
$$ 0 \rightarrow H^0(\mathbb{R}^2)=\mathbb{R} \rightarrow H^0(M_1)\oplus H^0(M_1) = \mathbb{R}^2 \rightarrow H^0(M_2) $$
$$ \rightarrow H^1(\mathbb{R}^2)=0 \rightarrow H^1(M_1)\oplus H^1(M_1) = \mathbb{R}^2 \rightarrow H^1(M_2) $$
$$ \rightarrow H^2(\mathbb{R}^2)=0\rightarrow H^2(M_1)\oplus H^2(M_1) = 0 \rightarrow H^2(M_2) $$
$$ \rightarrow 0 \cdots $$
Where I have used the poincare lemma in the left column, and that $S^1\sim M_1$ in the middle column. This will show that $h^2(M_d) = 0$ by inducting.