Can anyone check my proof that $H^1(\Sigma-\{p\})=0$ for a compact and orientable surface $\Sigma$?

Can anyone check my proof that $H^1(\Sigma-\{p\})=0$ for a compact and
orientable surface $\Sigma$?

I have the following problem: Let $\Sigma$ be a compact and orientable
surface. Show that $H^1(\Sigma-\{p\})=0$ for every $p\in \Sigma$.
Can anyone check my proof and give suggestions?
Sketch of proof: Let $B$ be a closed ball centered in $p\in \Sigma-\{p\}$
and consider the isomorphism $$\phi:H^1(\mathbb S^1)\rightarrow \mathbb
R,\ [\omega]\mapsto \int_{\mathbb S^1}\omega.$$ If $\imath:\partial
B\hookrightarrow \Sigma-\{p\}$ is the inclusion then
$\imath^*:H^1(\Sigma-\{p\})\rightarrow H^1(\partial B)$ is symply the
restriction. Since $H^1(\partial B)\simeq H^1(\mathbb S^1)$ we might
consider the composition $$\phi\circ \imath^*:H^1(\Sigma-\{p\})\rightarrow
\mathbb R.$$ Notice $\phi\circ \imath^*$ is given by, $$\phi\circ
\imath^*([\omega])=\phi(\imath^*[\omega])=\phi([\imath^*\omega])=\int_{\mathbb
S^1}\imath^*\omega=\int_{\partial B}\imath^*\omega.$$ I'm not pretty sure
if I really can write the above equality.. Hence, by Stokes theorem,
$$\int_{\partial
B}\imath^*\omega=\int_{\partial(M-B^\circ)}\imath^*\omega=\int_{M-B^\circ}d\omega=0,$$
(here $B^\circ$ is the interior of $B$) for $d\omega=0$. Therefore
$\phi\circ \imath^*([\omega])=0$ for all $[\omega]\in H^1(\Sigma-\{p\})$,
so $\phi\circ \imath^*=0$ and since $\phi$ is invertible $\imath^*=0$ so
that $\textrm{ker}(\imath^*)=H^1(\Sigma-\{p\})$. Since $\imath^*$ is
injective $H^1(\Sigma-\{p\})=0$ (is $\imath^*$ really injective?).