Question about maps of $S^{3}$-bundles Consider an $S^3$-bundle over a closed surface $\Sigma$ with genus $g>0$, and call the total space $X$. Suppose there is a smooth map $F: X  \rightarrow \Sigma$ such that $F$ induces an isomorphism $F_{*} : H_{2}(X,\mathbb{Z}) \rightarrow H_{2}(\Sigma,\mathbb{Z})$.

Question Can we always homotope $F$ so that it is constant on the fibers of the $S^{3}$-bundle.

 A: Here is an argument that is essentially the same as Oscar's, but organized a little differently.
First, we have a fibration $S^3\to X\to\Sigma$, giving an exact sequence 
$$ \pi_2(\Sigma) \to \pi_1(S^3)=0 \to \pi_1(X)\xrightarrow{\pi_*}\pi_1(\Sigma) \to \pi_0(S^3) = 0, $$
which proves that $\pi_*$ is an isomorphism.
Next put $G=\pi_1(\Sigma)$ and let $U$ be the universal cover of $\Sigma$, so that $\Sigma=U/G$.  For any connected based space $Z$, there is a natural map 
$$ \alpha_Z\colon [Z,\Sigma] \to \text{Hom}(\pi_1(Z),G)/\text{conjugacy}. $$
(We have to take the conjugacy quotient because we are considering unbased homotopy classes of unbased maps.)  It is standard that $U$ is homeomorphic to $\mathbb{R}^2$ and so is contractible.  Using this, covering theory proves that $\alpha_Z$ is always bijective.  By taking $Z=\Sigma$ or $Z=X$, we deduce that the map 
$$\pi^*\colon[\Sigma,\Sigma]\to[X,\Sigma]$$ 
is bijective.  It follows that every map $X\to\Sigma$ is homotopic to one that factors through $\pi$, as required.
A: Yes, and you do not need the hypothesis on $H_2$. The projection map $\pi: X \to \Sigma$, having fibre $S^3$, is 3-connected, and hence we can attach cells of dimension $\geq 4$ to $X$ to form a CW-complex $X'$, and extend the map $\pi$ to a homotopy equivalence $\pi' : X' \to \Sigma$, which then has a homotopy inverse $s : \Sigma \to X'$.
On the other hand, as $\Sigma$ has trivial homotopy groups in degree $\geq 2$, the map $F : X \to \Sigma$ may be extended to $X'$, giving a factorisation $F : X \to X' \overset{F'}\to \Sigma$. The composition
$$\hat{F} : X \overset{\pi}\to \Sigma \overset{s}\to X' \overset{F'}\to \Sigma$$
is homotopic to $F$, as $s \circ \pi$ is homotopic to the inclusion (because $s$ is a homotopy inverse to $\pi'$). But $\hat{F}$ is constant on fibres of $\pi$, because $\pi$ is the first map in the composition.
