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For a locally compact space, compactly supported cohomology iswith transition maps induced by enlarging supports. In ${\mathbb{R}}^n$ closed balls are cofinal; their complements retract onto $S^{n-1}$. The relative cohomology long exact sequence and contractibility of ${\mathbb{R}}^n$ giveThis is compactly supported cohomology of Euclidean space; for $n=0$ it is simply the cohomology of a point.

For an oriented $n$-manifold without boundary, Poincare duality in its compact-support form states that capping with the locally finite fundamental class gives $H_c^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$. If $M$ is closed, the ordinary fundamental class $[M]\in H_n(M;\mathbb{Z})$ gives $H^q(M;\mathbb{Z})\cong H_{n-q}(M;\mathbb{Z})$ and a perfect complementary-degree pairing over $\mathbb{Q}$.

Choose an oriented embedded closed ball $B$ in a connected closed oriented $M$. Collapse $M\setminus \mathrm{int}B$ to a point, identifying $B/\partial B$ with $S^n$ using its orientation. The quotient map is continuous. By excision, the fundamental class maps to the local oriented generator in $H_n(B,\partial B)$, and then to the generator of $H_n(S^n)$. Hence the collapse map has $\boxed{\mathrm{deg}f=1}$.

The reverse direction is not always possible. More generally a degree-one map $f:N\to M$ between closed oriented manifolds induces an injective map on rational cohomology: if $\alpha \ne 0$, duality supplies $\beta$ with $⟨\alpha ⌣\beta ,[M]⟩\ne 0$, andIn particular there is no degree-one map $S^2\to T^2$, since $H^1(T^2;\mathbb{Q})={\mathbb{Q}}^2$ but $H^1(S^2;\mathbb{Q})=0$. This is a counterexample to an unconditional reverse assertion.

Finally let $p:{\mathbb{CP}}^3\to S^4$ be the given bundle projection. The standard cohomology ring of complex projective space is $H^{\ast }({\mathbb{CP}}^3;\mathbb{Z})=\mathbb{Z}[h]/(h^4)$, $|h|=2$; the integral generator $u$ of $H^4(S^4)$ has $p^{\ast }u=m{h}^2$ for some integer $m$. If a section of a fiber bundle $s$ existed, $p\circ s=1$ would give $s^{\ast }{p}^{\ast }=1$ by functoriality. But $H^2(S^4)=0$, so $s^{\ast }h=0$ and $s^{\ast }(m{h}^2)=0$, contradicting $s^{\ast }{p}^{\ast }u=u\ne 0$. The facts used here are the sphere cohomology groups, the projective-space ring just stated, and multiplicative functoriality of cohomology. Thus the bundle has no section, the cohomological obstruction to a section of the twistor bundle.