deRham cohomology of $S^n$ without Mayer-Vietoris I've got a few days left at the end of a differential geometry class, and would like to compute the deRham cohomology of $S^n$.  We've just proved the Poincare lemma, so I know the cohomology of $R^n$, and homotopy invariance of cohomology is an easy consequence.
The way to compute $H^*(S^n)$ is with Mayer-Vietoris, for example as in Bott and Tu.  But I don't have time to fully develop the cohomological algebra for that, just to then apply it in only one case.  I'd rather get the computation for $S^n$ directly and use it for a couple of applications.
So, it seems to me that one could trace through the Mayer-Vietoris argument in this specific case. At least I'd like to show that a closed $k$-form on $S^n$ is exact when $0 < k < n$.
Here's how I think the argument could go:
Let $\omega$ be a $k$-form on $S^n$.
Let $U$ and $V$ be $S^n$ minus its north and south pole, respectively.  Then homotopy invariance and the Poincare lemma give $k-1$ forms $\alpha$ on $U$ and $\beta$ on $V$ with $d\alpha = \omega$ and $d\beta = \omega$.
Now use a partition of unity $f$ and set $\gamma = f_U \alpha + f_V \beta$, a $k-1$ form on $S^n$.  So $\omega - d\gamma$ is now supported on $U \cap V$.  Since $U\cap V$ is homotopic to $S^{n-1}$, induction gives that $\omega - d\gamma$ is exact, so $\omega = d\tau + d\gamma$.
The problem is that $\omega - d\gamma$ is exact when restricted to $U \cap V$, and I don't see how why $\tau$ would extend to a form on $S^n$.
Am I missing something?  Is there a better approach entirely?
 A: Suppose $1 < k < n$.
Let $\omega$ be a $k$-form on $S^n$.
Let $U$ and $V$ be $S^n$ minus its north and south pole, respectively.  Then homotopy invariance and the Poincare lemma give $k-1$ forms $\alpha$ on $U$ and $\beta$ on $V$ with $d\alpha = \omega$ and $d\beta = \omega$.
On $U \cap V$, $d(\alpha -\beta) = 0$.  Since $H^{k-1}(S^{n-1}) = 0$, there is a $k-2$ form $\tau$ on $U \cap V$ with
$d\tau = (\alpha - \beta) |_{U\cap V}$.
Now let $f_U, f_V$ be a partition of unity subordinate to $U,V$.  Then on $U\cap V$, $d(f_U\tau) + d(f_V\tau) = \alpha - \beta$.
Since $f_U\tau$ extends (by 0) to a form on $V$, $\beta + d(f_U \tau)$ is defined on $V$.
Since $f_V\tau$ extends to a form on $U$, $\alpha - d(f_V \tau)$ is defined on $U$.
Put $\sigma = \beta + d(f_U\tau) = \alpha - d(f_V\tau)$.  $\sigma$ is defined on all of $S^n$, and $d\sigma = \omega$, so that $\omega$ is exact.
In the $k=1$ case, $\alpha$ and $\beta$ differ by a constant, so they extend to all of $S^n$ and $\omega = d\alpha$.
When $k=n$, show that any $n$ form with integral 0 is exact, so $H^n(S^n)$ is generated 
by the volume form.  The only change from the $k < n$ case is that one needs to check that the 
integral of $\alpha-\beta$ is zero over $S^{n-1}$. Apply Stokes' theorem  to the lower hemisphere (for $\alpha$) and the upper hemisphere (for $\beta$), so that
$\int_{S^{n-1}} \alpha - \beta = \int_{S^n}\omega = 0$.
A: Have you done any integration theory?  (I assume you have, otherwise you wouldn't necessarily know what the deRham cohomology does for you.)  The fastest proof I know is: 


*

*Take a closed $k$-form $\omega$ on $S^n$, note that $g^\ast\omega$ is cohomologous to $\omega$ for all $g\in \mathrm{SO}(n{+}1)$ (since $\mathrm{SO}(n{+}1)$ is connected.  

*Conclude that $\omega$ is cohomologous to $\bar\omega$, the average over $\mathrm{SO}(n{+}1)$ of $g^\ast\omega$ as $g$ varies over $\mathrm{SO}(n{+}1)$.  

*But $\bar\omega$ is invariant under the action of $\mathrm{SO}(n{+}1)$, so its value at $x\in S^n$ must be invariant under the subgroup (isomorphic to $\mathrm{SO}(n)$) that stabilizes $x$.  

*However, $\mathrm{SO}(n)$ acting on $\mathbb{R}^n$ only fixes nonzero forms in degree $0$ and $n$.  

*Thus, if $\bar\omega$ is not zero, it must be either a constant function ($k=0$) or a multiple of the volume form ($k=n$).
A: For any connected and oriented $n$-manifold $M$, the sequence:
$$\Omega_c^{n-1}(M) \to \Omega_c^n(M) \to \mathbb{R} \to 0$$
is exact, where the first map is $d$ and the second map is $\int_M$.  A detailed proof can be found in Madsen and Tornhave's "From Calculus to Cohomology", but here's a sketch:


*

*Prove the result for $M = \mathbb{R}^n$.  Since every top degree form on $\mathbb{R}^n$ is a multiple of the volume form, this comes down to proving that any smooth compactly supported function $f$ on $\mathbb{R}^n$ with total integral $0$ can be written as $\sum_{j=1}^n \frac{\partial f_j}{\partial x_j}$ for some smooth functions $f_j$, and this can be done with a bit of calculus.

*For any $\omega \in \Omega_c^n(M)$ and any coordinate neighborhood $U \subseteq M$ there is some $\eta \in \Omega_c^{n-1}(M)$ such that $\omega - d\eta$ is supported in $U$.  Indeed, just pick any $\omega_0$ compactly supported in $U$ with total integral $1$ and deduce from step $1$ that the form $\omega - (\int_M \omega) \omega_0$ is exact.

*Let $\omega \in \Omega_c^n(M)$ be a form with total integral $0$ and let $U$ be a coordinate neighborhood in $M$.  By step 2 there is a form $\eta \in \Omega_c^{n-1}(M)$ such that $\omega - d\eta$ is supported in $U$, and this form has total integral $0$ by Stokes' theorem.  But by step 1 $(\omega - d\eta)|_U = d\xi$ for some $\xi \in \Omega_c^{n-1}(U)$; extending $\xi$ by $0$ to all of $M$, we obtain $\omega = d(\eta + \xi)$.


An immediate corollary is that $H^n(M) \cong \mathbb{R}$ (with the isomorphism given by integration) for any closed oriented $n$-manifold $M$.  
The simplest proof I can think of that $H^k(S^n)$ vanishes for $0 < k < n$ uses Poincare duality.  Any closed $k$-form $\omega$ cohomologous to a closed $k$-form $\omega_0$ supported in a small neighborhood of its Poincare dual.  For $k < n$ the Poincare dual is a proper subset of $S^n$ and thus $\omega_0$ can be viewed as a closed (compactly supported) $k$-form on $\mathbb{R}^n$ via stereographic projection.  Thus $\omega_0$ - and hence $\omega$ - is exact since you already calculated the de Rham cohomology of $\mathbb{R}^n$.  Of course, this argument only helps if you have covered Poincare duality...
