simply connected fibre bundle Suppose that $B,F$ are closed manifolds. Suppose $\pi_{1}(B)=\{1\}$. Suppose furthermore that there is a surjective homomorphism $H: \pi_{2}(B) \rightarrow \pi_{1}(F)$. Does there exist an orientable $F$-bundle over $B$, denoted by $E$, with $\pi_{1}(E) = \{1\}$?
(Notation: Orientable means that the transition functions are in the orientation preserving homeomorphisms of $F$. By closed manifold I mean connected compact oriented.)
 A: Not neccesarily.
Let $F = (S^1\times S^3) \sharp \mathbb{C}P^2$ and let $B = S^2$.  Note that by van Kampen, $\pi_1(F) = \pi_1(S^1) = \mathbb{Z}$ and $\pi_2(S^2) = \mathbb{Z}$, so of course there is a surjective map $\pi_2(B)\rightarrow \pi_1(F)$.
Now, suppose we have a bundle $F\rightarrow E\rightarrow B$.
Proposition  The space $E$ cannot be simply connected.  In fact, $E$ cannot have finite fundamental group.
Proof:  We will use the Serre spectral sequence.
To that end, note $H^k(F) = \mathbb{Z}$ for $0\leq k\leq 4$, and is $0$ otherwise.  Let $x_k\in H^k(F)$ be a generator.  Then the cohomology ring structure is determined by graded commutativity together with the fact that $x_1 x_2 = x_3x_2 = 0$, and that $x_2^2 = \pm x_1 x_3 = \pm x_4$.  Let $y\in H^2(S^2)$ denote a generator.
Consider the Serre spectral sequence for the bundle.  Note that the $E_3$ pages is the $E_\infty$ page because all differentials on $E_r$, $r\geq 3$ page vanish for trivial reasons.
Now, the differential $d:\langle 1\otimes x_2\rangle = E_2^{0,2}\rightarrow E_2^{2,3} = \langle y\otimes x_3\rangle$ has the form $d(1\otimes x_2) = \lambda (y\otimes x_1)$ for some $\lambda \in \mathbb{Z}$.
The Leibniz formula now gives $$d((1\otimes x_2)^2) = d(1\otimes x_2)(1\otimes x_2) + (1\otimes x_2) d(1\otimes x_2) = 2(1\otimes x_2) d((1\otimes x_2),$$ because $x_2$ has even degree.  This now gives $$d(x_2^2) = 2\lambda (1\otimes x_2) (y\otimes x_1) = 2\lambda (y\otimes x_2x_1) = 0$$ since $x_2 x_1= 0$.
Since $x_2^2$ is a generator of $H^4(F)$, it now follows that the differential $d:E_2^{0,4}\rightarrow E_2^{2,3}$ is trivial.  Thus, $E_\infty^{2,3}\cong \mathbb{Z}$, which now implies $H^5(E) = \mathbb{Z}$.
Now, the bundle is orientable because $S^2$ is simply connected.  Since $F$ and $B$ are orientable, this now implies $E$ is orientable, so we can use Poincare duality.  This now gives that $H_1(E) = \mathbb{Z}$.  In particular, $\pi_1(E)$ is infinite.  $\square$
If you look at the long exact sequence in homotopy groups associated to the bundle $F\rightarrow E\rightarrow B$, the fact that $\pi_1(E)$ is infinite is equivalent to the assertation that the map $\pi_2(B)\rightarrow \pi_1(F)$ is trivial.  So, we have proved the following regarding the connecting homomorphism.
Proposition  For $F = (S^1\times S^3)\sharp \mathbb{C}P^2$ and $B = S^2$, in any fiber bundle $F\rightarrow E\rightarrow B$, the connecting homomophism $\pi_2(B)\rightarrow \pi_1(F)$ is the trivial map. 
