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.