Relative cohomology versus cohomology.

Question

Let $S$ be a closed oriented surface, $X$ a finite set of points on $S$. Is it true that $$ H^1(S \setminus X, \mathbb{C}) \simeq H^1(S,X,\mathbb{C}) $$

Answer 1

This seems to be true. Let me answer the question in homology. The universal coefficient theorem in Cohomology should then transfer the result.

We have

$$H_1(X) \rightarrow H_1(S)\rightarrow H_1(S,X)\rightarrow \tilde{H}_0(X)\rightarrow \tilde{H}_0(S)$$ which is

$$0 \rightarrow \bigoplus_{g} \mathbb{Z}\rightarrow H_1(S,X)\rightarrow \bigoplus_{|X|-1} \mathbb{Z}\rightarrow 0$$ ($g$ is the genius of the surface) So $H_1(S,X)$ should be a free abelian group of rank $g+|X|-1$.

On the other hand $H_1(S\setminus X)$ should have the same rank an before, we show this using Mayer-Vietorius let $A=S\setminus X$ and $B$ the disjoint union of disks about so the points of $X$, so $A\cap B$ the disjoint union of $|X|$ circles,

$$H_2(A)\oplus H_2(B)\rightarrow H_2(S)\rightarrow H_1(A\cap B) \rightarrow H_1(A)\oplus H_1(B)\rightarrow H_1(S)\rightarrow \tilde{H}_0(A\cap B) \rightarrow \tilde{H}_0(A)\oplus \tilde{H}_0(B)\rightarrow \tilde{H}_0(S)$$ which gives,

$$0\rightarrow \mathbb{Z}\rightarrow \bigoplus_{|X|}\mathbb{Z} \rightarrow H_1(A)\rightarrow \bigoplus_{g}\mathbb{Z}\rightarrow \bigoplus_{|X|-1}\mathbb{Z}\rightarrow \bigoplus_{|X|-1}\mathbb{Z}\rightarrow 0$$ And again a rank calculation gives that $H_1(A)=H_1(S\setminus X)$ has rank $g+|X|-1$.