= Solution
For a locally compact space, <compactly supported cohomology> is
$$
H_c^q(X;\mathbb Z)=\varinjlim_{K\subset X\text{ compact}}H^q(X,X\setminus K;\mathbb Z),
$$
with 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 $\mathbb R^n$ give
$$
\boxed{H_c^q(\mathbb R^n;\mathbb Z)=\begin{cases}\mathbb Z,&q=n,\\0,&q\ne n.\end{cases}}
$$
This 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 $M\setminus\operatorname{int}B$ 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 \b[the collapse map has $\boxed{\deg f=1}$].
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 $\alpha\ne0$, duality supplies $\beta$ with $\langle\alpha\smile\beta,[M]\rangle\ne0$, and
$$
\langle f^*\alpha\smile f^*\beta,[N]\rangle
=\deg(f)\langle\alpha\smile\beta,[M]\rangle\ne0.
$$
In particular \b[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 $H^*(\mathbb{CP}^3;\mathbb Z)=\mathbb Z[h]/(h^4)$, $|h|=2$; the integral generator $u$ of $H^4(S^4)$ has $p^*u=mh^2$ for some integer $m$. If a <section of a fiber bundle> $s$ existed, $p\circ s=1$ would give $s^*p^*=1$ by functoriality. But $H^2(S^4)=0$, so $s^*h=0$ and $s^*(mh^2)=0$, contradicting $s^*p^*u=u\ne0$. The facts used here are the sphere <cohomology> groups, the projective-space ring just stated, and multiplicative functoriality of <cohomology>. Thus \b[the bundle has no section], the <cohomological obstruction to a section of the twistor bundle>.
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