Prove $H^{k}_\text{deRham} (M \times S^{n}) \cong H^{k}_\text{deRham} (M) \bigoplus H^{k-n}_\text{deRham} (M)$ using Mayer-Vietoris I want to show $H^{k}_\text{deRham} (M \times S^{n}) \cong H^{k}_\text{deRham} (M) \bigoplus H^{k-n}_\text{deRham} (M)$ using Mayer-Vietoris only, that is without using Kunneth or anything similar. Using the obvious decomposition $S^{n} = S^{n} - \{(0,..,1)\} \cup S^{n} - \{(0,...,-1) \}$ and induction on $n$ I couldn't manage to get very far by using the exactness of the corresponding Mayer-Vietoris sequence. I remember that in Hatcher a similar exercise for singular homology was proved by passing through to relative homology, so im assuming there is something extra that im missing. Any hints would be greatly appreciated!
Edit: Solution moved to answers.
 A: We prove the claim by induction on $n$. The case $n=0$ is trivial since $S^{0}$ is two points, so $M \times S^{0} \cong M \sqcup M$, so the claim follows from additivity of cohomology under $\sqcup$.
Proceed by induction. Write $S^{n} = S^{n}_{+} \cup S^{n}_{-}$ where $S^{n}_{+}$ is $S^{n}$ with the north pole removed, and $S^{n}_{-}$ is $S^{n}$ with the south pole removed.
Our induction hypothesis is as follows: We will have in our LES a map $H^{k} (M \times S^{n}_{+})  \bigoplus H^{k} (M \times S^{n}_{-} ) \to H^{k} (M \times (S^{n}_{+} \cap S^{n}_{-} )$. Identify using homotopy equivalences $M \times S^{n}_{+} \cong M$, $M \times (S^{n}_{+} \cap S^{n}_{-} ) \cong  M \times S^{n-1} \times \mathbb{R} \approx M \times S^{n-1}$. Our induction hypothesis is now that the map $H^{k} (M) \bigoplus H^{k} (M) \to H^{k} (M \times S^{n-1} ) = H^{k} (M) \bigoplus H^{k - n + 1} (M)$ (which is given by passing by using a homotopy equivalence)  satisfies $Im = H^{k} (M)$ (As in, the left summand), and $Ker = \Delta = \{ (v,v) | v \in H^{k}(M) \}$.
Define $S^{n}_{+-} = S^{n}_{+} \cap S^{n}_{-}$. Our Mayer-Vietoris sequence is:
\begin{align*}
    ... \to H^{k-1} (M \times S^{n}_{+-} ) \to H^{k} (M\times S^{n}) \to H^{k} (M\times S^{n}_{+} ) \bigoplus H^{k}(M\times S^{n}_{-}) \to H^{k} (M\times S^{n}_{+-}) \to ... 
\end{align*}
Which, by the induction hypothesis, and by the lemma from the beginning, is:
\begin{align*}
    ... \to H^{k-1}(M)\bigoplus H^{k-n}(M) \to H^{k} (M \times S^{n}) \to H^{k}(M)\bigoplus H^{k}(M) \to H^{k}(M)\bigoplus H^{k-n+1}(M) \to ...
\end{align*}
Using exactness of the sequence and the induction hypothesis, we extract the short exact sequence:
\begin{align*}
    0\to H^{k-n}(M) \to H^{k}(M \times S^{n} ) \to \Delta \to 0
\end{align*}
Identifying $H^{k-n} (M)$ with its' image in $H^{k}(M\times S^{n})$, $\Delta$ with $H^{k}(M)$ and using the exactness of the sequence + the linear algebra fact $V \cong V/W \bigoplus W$ for any subspace $W \leq V$ (that may require axiom of choice in the infinite dimensional case), we finally deduce:
\begin{align*}
    H^{k} (M\times S^{n} ) \cong H^{k} (M) \bigoplus H^{k-n} (M)
\end{align*}
All that's left is showing the induction hypothesis still holds. The fact that the short exact sequence leads to an isomorphism with the direct sum follows from the axiom of choice, but we can choose it to "work well" with the earlier needed map $H^{k} (M) \bigoplus H^{k} (M) \to H^{k} (M \times S^{n} )$. Specifically, choose any complementary subspace $W$ of $H^{k}(M\times S^{n} )$ to $Ker(g)$, so that $Ker(g) \bigoplus W = H^{k} (M\times S^{n} )$ and $g$ is consequentially an isomorphism between $W$ and $\Delta$ (finding such $W$ requires Zorn's lemma). Under this choice, $H^{k} (M) \bigoplus H^{k} (M) \to H^{k} (M) \bigoplus H^{k-n} (M)$ is written as $(v,w)\mapsto(v-w,0)$, and so it indeed satisfies both requirements of the induction hypothesis.
