# 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?).

• A punctured compact surface deformation-retracts to a wedge of circles, so its $H^1$ is rarely zero. – Mariano Suárez-Álvarez Aug 9 '13 at 20:26
• @MarianoSuárez-Alvarez you mean the result is not true? – PtF Aug 9 '13 at 20:29
• Indeed. Take for example $\Sigma$ to be a torus: it defirmation-retracts onto the union of one of its equators and one of its parallels. – Mariano Suárez-Álvarez Aug 9 '13 at 20:36
• Notice that the fact that this is not zero follows immediately from the answer you got here! math.stackexchange.com/questions/462129/… – Mariano Suárez-Álvarez Aug 9 '13 at 20:58

Consider the one-form $dx$ on the punctured torus $T^{2*}=(\mathbb{R}^2 - \mathbb{Z}^2) / \mathbb{Z}^2$. Note that the image of the line $\gamma(t) = (t,\frac{1}{2})$ under the quotient is a circle, hence is a $1$-chain, and $$\int_{\gamma}dx = 1.$$ This implies that $dx$ is not a coboundary, for if it were and $dx = df$ for some function $f$, then by Stokes' theorem, $\int_\gamma df = f(\gamma(1)) - f(\gamma(0)) = 0$ since $\gamma(1) = \gamma(0)$.
Therefore $[dx]\in H^1_{dR}(T^{2*})$ is a nontrivial cohomology class.
To your proof, why should we believe $i^*$ is injective? Consider again $T^2=\mathbb{R}^2/\mathbb{Z}^2$. Let $\gamma$ be as above and $\delta = (\frac{1}{2},t)$. Now $dx(\gamma) = 1$, but $dx(\delta) = 0$; $dy(\gamma) = 0$ while $dy(\delta) = 1$. If $B$ is centered at $(0,0)$, then $dx(\partial B) = dy(\partial B) = 1 + 0 - 1 - 0 = 0$. This is an example of why a non-injective $i^*$ unless I am very much off my game today.
• «is not a boundary» $\leadsto$ «is not exact». – Mariano Suárez-Álvarez Aug 9 '13 at 21:33
• :-) ${}{}{}{}{}$ – Mariano Suárez-Álvarez Aug 9 '13 at 21:39