De rham cohomology of spheres

For $n\ge 1$, write points as $(x\mathrm{\prime },z)$ and remove the poles to form $U,V$. Stereographic coordinates $x\mathrm{\prime }\mathrm{/}(1-z)$ and $x\mathrm{\prime }\mathrm{/}(1+z)$ identify these opens with ${\mathbb{R}}^n$; the first inverse is $y\mapsto (2y\mathrm{/}(1+\mathrm{\mid }y{\mathrm{\mid }}^2),(\mathrm{\mid }y{\mathrm{\mid }}^2-1)\mathrm{/}(1+\mathrm{\mid }y{\mathrm{\mid }}^2))$, and changing the sign of the last coordinate gives the second. Substitution verifies both inverses. Positive-degree cohomology of each open vanishes by Poincaré, and its degree-zero group is $\mathbb{R}$.

The overlap is diffeomorphic to $S^{n-1}\times \mathbb{R}$ by $(x\mathrm{\prime },z)\mapsto (x\mathrm{\prime }\mathrm{/}\mathrm{\mid }x\mathrm{\prime }\mathrm{\mid },z\mathrm{/}\mathrm{\mid }x\mathrm{\prime }\mathrm{\mid })$, with inverse $(u,t)\mapsto (u\mathrm{/}\sqrt{1+t^2},t\mathrm{/}\sqrt{1+t^2})$. The homotopy $(u,t,s)\mapsto (u,(1-s)t)$ retracts it smoothly onto $S^{n-1}$. For $n=1$ the overlap is two contractible components, so its $H^0$ is ${\mathbb{R}}^2$ and its positive groups vanish. The map on $H^0$ is $(a,b)\mapsto (b-a,b-a)$; its kernel is the diagonal and its cokernel is $\mathbb{R}$, via $(c,d)\mapsto d-c$. Exactness therefore gives $H^0(S^1)=H^1(S^1)=\mathbb{R}$. In degrees $k\ge 2$ the form spaces on this one-manifold vanish, so its cohomology also vanishes.

For $n\ge 2$ the overlap is connected: $S^{n-1}$ is path connected, since non-antipodal points join by normalized line segments and antipodal points join through one perpendicular unit vector. Thus the degree-zero difference map is the surjection $(a,b)\mapsto b-a$ onto $\mathbb{R}$. Exactness gives $H^0(S^n)=\mathbb{R}$ and $H^1(S^n)=0$. For $k\ge 2$ both adjacent positive-degree groups of $U,V$ vanish, so exactness gives $H^k(S^n)\cong H^{k-1}(S^{n-1})$. Repeatedly applying this identity reaches the circle calculation or degree one, proving all asserted positive degrees. Negative degrees vanish by the complex convention. Finally $S^0$ is two points, each with only ${\mathrm{\Omega }}^0=\mathbb{R}$, and finite disjoint union gives its stated groups.