TITLE: deRham cohomology of $S^n$ without Mayer-Vietoris
QUESTION [16 upvotes]: 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?

REPLY [6 votes]: 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
$\int_{S^{n-1}} \alpha - \beta = \int_{S^n}\omega = 0$.