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 $F\_{\*} : H\_{2}(X,\mathbb{Z}) \rightarrow H\_{2}(\Sigma,\mathbb{Z})$.



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


