De Rahm cohomology of a sphere, help with proof I am working through Guillemin and Pollack's proof that the de Rahm cohomology of the sphere is $H^p(\mathbf{S}^k) = \mathbf{R}$ for $p = 0$ and $p = k$ and $H^p(\mathbf{S}^k) = 0$ otherwise.  Here, $k > 0$.  
I am having difficulty understanding one of the maps involved in the proof.  
The proof is by induction on $k$.  Assume we are in the inductive step, and we wish to prove the result for $\mathbf{S}^k$, $k > 1$.  We use a Mayer-Vietoris type of argument.  Decompose $\mathbf{S}^k$ as $U_1 \cup U_2$, where $U_1$ is the sphere minus the south pole $S$ and $U_2$ is the sphere minus the north pole $N$.  I am in the middle of showing the following result:

Proposition:  For $p > 1$, $H^{p-1}(U_1 \cap U_2) \cong H^p(U_1 \cup U_2)$.  

The proof goes by constructing explicit maps, each of which is the other's inverse.  I am having trouble understanding why the specific map $H^{p-1}(U_1 \cap U_2) \to H^p(U_1 \cup U_2)$ is induced.  
Specifically:  Let $Z^p(M)$ denote the closed $p$-forms on a manifold $M$, $B^p(M)$ the exact $p$-forms on the manifold, and $H^p(M) = Z^p(M)/B^p(M)$ the $p^{th}$ cohomology on the manifold.  
Define a map $G \colon Z^{p-1}(U_1 \cap U_2) \to H^p(U_1 \cup U_2)$ as follows.  Let $\nu$ be a closed $(p-1)$-form on $U_1 \cap U_2$.  Note that $\nu$ may blow up at either $N$ or $S$.  Choose functions $\rho_1, \rho_2 \colon U_1 \cup U_2 \to \mathbf{R}$ such that $\rho_1$ vanishes on a neighborhood of $N$, $\rho_2$ vanishes on a neighborhood of $S$, and $\rho_1 + \rho_2 = 1$.  Define a $(p-1)$-form $\nu_1$ on $U_1$ by $\nu_1 = \rho_1 \nu$ and define a $(p-1)$-form $\nu_2$ on $U_2$ by $\nu_2 = -\rho_2 \nu$.  Define the $p$-form $\omega$ on $U_1 \cup U_2$ by setting $\omega = d \nu_1$ on $U_1$ and $\omega = d \nu_2$ on $U_2$.  Then $\omega$ is well-defined since on $U_1 \cap U_2$, $d \nu_1 - d \nu_2 = d \nu = 0$.  Clearly, $\omega$ is closed.  Lastly, set $G(\nu)$ to be the class of $\omega$ in $H^p(U_1 \cup U_2)$.  I was able to prove that $G(\nu)$ is well-defined in the sense that it is independent of the chosen partition of unity.  

Question:  Why is it the case that $G$ induces a map $H^{p-1}(U_1 \cap U_2) \to H^p(U_1 \cup U_2)$.  That is, why is it true that if $\nu$ is closed and exact, then $\omega$ is also exact?  

I realized after working on this for a bit that my method couldn't (shouldn't?) work because it would prove that $\omega$ is exact for any closed $\nu$.  Specifically, I was only going off the fact that $d \nu = 0$.  My natural instinct now is to show $\nu_1$ extends to $S$ (or $\nu_2$ extends to $N$), but why would $\nu$ being exact mean such an extension exists?  That doesn't seem right to me...  (Just to reiterate, $\omega$ is very close to being exact already; $\omega = d \nu_1$ (or $= d \nu_2$) except at just one point at which $\nu_1$ is not, a priori, defined.)  
Hope someone can help clear this up for me, understanding these maps has been very frustrating!
Note: It might be helpful that $U_1$ and $U_2$ are contractible (hence have vanishing cohomology), $U_1 \cap U_2 \cong \mathbf{R} \times \mathbf{S}^{k-1}$, $H^p(\mathbf{R} \times \mathbf{S}^{k-1}) \cong H^p(\mathbf{S}^{k-1})$, we are in the induction step, etc... ???
 A: This is the general construction of a Mayer-Vitoris argument. I did not read that book, but one can show that one has the following exact sequence (in general for any open set $U_1, U_2$ in $M$ so that $U_1 \cup U_2 = M$)
$$\begin{split}
\cdots H^p(M) \to H^p(U_1)& \oplus H^p(U_2) \to H^p(U_1\cap U_2) \overset{J}{\to} \\
&H^{p+1}(M) \to H^{p+1}(U_1)\oplus H^{p+1}(U_2) \cdots
\end{split}$$
you are dealing with the map $J$. You have already 
$$ \tilde J : Z^p (U_1\cap U_2) \to H^{p+1} (M),$$
where 
$$\tilde J(\omega) = \begin{cases} d(\rho_1 \omega) &\text{on } U_1, \\-d(\rho_2 \omega )&\text{on } U_2.\end{cases}$$
to show that $\tilde J$ descends to a map on the cohomology, let $\omega = d\eta$ for some $\eta \in H^{p-1} (U_1 \cap U_2)$. Then 
$$\tilde J (d\eta)= \begin{cases} d(\rho_1 d\eta) = d(-d\rho_1 \wedge \eta) &\text{on } U_1, \\-d(\rho_2 d\eta ) = d(d\rho_2\wedge \eta)&\text{on } U_2.\end{cases}$$
but $\rho_1+\rho_2 = 1$ implies $d\rho_1 = -d\rho_2$ on $U_1\cap U_2$. Thus 
$$\nu =\begin{cases}  -d\rho_1 \wedge \eta &\text{on } U_1, \\d\rho_2\wedge \eta&\text{on } U_2.\end{cases}$$
is a well-defined $p$-forms on $M$ so that $d\nu = \tilde J(d\eta)$. Thus $[\tilde J(d\eta) ] =0$ and so $\tilde J$ descends to a map on cohomology. 
