Let $F \to E \to B$ be a fiber bundle, and assume $E$ is connected and $F$ is simply connected. If the base space $B$ is simply connected, then is the total space $E$ also simply connected?
I think this is true by the homotopy lifting property, but I am not positive. [My apologies if this question is silly; I'm just starting to learn differential/algebraic topology.]
EDIT: Edited to include the assumption that the fiber $F$ is simply connected.
The answer is yes. As @MoisheKohan pointed out, the tool to use here is the long exact sequence of a fiber bundle. I will assume that $F$ is path-connected and simply connected ($\pi_0(F) = 1, \pi_1(F) = 1$).
In this case, we have the (short) exact sequence $$1 = \pi_1(F) \stackrel{f}{\to} \pi_1(E) \stackrel{g}{\to} \pi_1(B) \stackrel{h}{\to} \pi_0(F) = 1$$ where $f, g, h$ are group homomorphisms. By exactness, $1 = im(f) = ker(g)$ (i.e., $g$ is injective) and $im(g) = ker(h) = \pi_1(B)$ (i.e., $g$ is surjective). Therefore, $\pi_1(E) \cong \pi_1(B)$. In particular, $B$ is simply connected if and only if $E$ is simply connected.