Cohomology of punctured surface

Question

Suppose $M$ is a compact oriented surface without boundary, $p$ is a point on $M$. Whether does the conclusion hold: $H^1(M-\{p\}) \cong R$ ? If yes, how to prove it (or please kindly point me to a reference)?

Answer 1

As Matt Samuel says as a comment, this is not true.

Take $M=S^2$, then $S^2-p\simeq D^2\simeq \ast$ (seen through a stereographic projection), so $\widetilde{H}^i(S^2-p; A)=0$ for all $i\geq0$ and all abelian groups $A$.

Other counter-examples include a genus $g$ surface $M_g$, which can be built as in page 51 of Hatcher's Algebraic Topology book (see https://www.math.cornell.edu/~hatcher/AT/AT.pdf), and we can then see that $$M_g-p\simeq \bigvee_{2g} S^1,$$ which is the wedge sum of $2g$-many 1-spheres. Then we immediately get $$H^1(M_g-p;A)\cong\bigoplus_{2g}A,$$ for any abelian group $A$.

In fact, by the classification of compact oriented surfaces, the result never holds for such spaces.