Morphism with connected fibers induce surjection on fundamental groups? Let $X,Y$ be path-connected finite CW complexes with base points $x_0,y_0$, let $f\colon X\to Y$ be a surjective continuous map, such that for every $y\in Y$, the fiber $f^{-1}(y)$ is path connected. In this case,  is the induced map $$f_*\colon\pi_1(X,x_0)\to\pi_1(Y,y_0)$$ on topological fundamental groups necessarily surjective? 
(If this is not true in general, will this be true in the case when $Y$ is a complex algebraic manifold, $X\subset Y\times\mathbb{P}^n$ a quasi-projective variety, and $f=\mathrm{pr}_1$?)
[I think one sufficient condition is that $f$ satisfies the "arc-lifting property": Sufficiently short arc $(-\epsilon,\epsilon)$ centered at any $y\in Y$ can be lifted to an arc in $X$. For then we can cover a path in $X$ by finitely many arcs in $X$, and join the arcs by path in the fibers. But I am not sure if this is always doable?]
 A: Assuming that your map $f\colon X \rightarrow Y$ is a map of CW complexes, the answer is yes.
In fact, you can get away with quite a bit less.  Assume that $X$ and $Y$ are arbitrary CW complexes equipped with basepoints $x_0 \in X^{(0)}$ and $y_0 \in Y^{(0)}$ and that $f\colon (X,x_0) \rightarrow (Y,y_0)$ is a map of CW complexes.  Furthermore, assume that for all vertices $v \in Y^{(0)}$, the preimage $f^{-1}(v)$ is connected and that for all $1$-simplices $e$ of $Y$, there is some $1$-simplex $E$ of $X$ that is taken to $e$ by $f$.  Then I claim that $f_{\ast}\colon \pi_1(X,x_0) \rightarrow \pi_1(Y,y_0)$ is surjective.
There is probably some fancy way of seeing this, but here's a down-to-earth argument.  Every element of $\pi_1(Y,y_0)$ can be represented by an edge path in the $1$-skeleton.  Let $e_1,e_2,\ldots,e_k$ be the edges traversed by that edge path.  The lift to $X$ is then as follows:


*

*Start at $x_0$.

*There is some edge $E_1$ of $X$ projecting to $e_1$; move in the fiber $f^{-1}(y_0)$ to the starting point of $E_1$ and then go across $E_1$.

*Letting $y_1$ be the ending point of $e_1$, there again exists some edge $E_2$ of $X$ projecting to $e_2$.  Move in the fiber $f^{-1}(y_1)$ to the starting point of $E_2$ and then go across $E_2$.

*etc.

*At the end of this process, you'll end up at a point of $f^{-1}(y_0)$.  Move in the fiber $f^{-1}(x_0)$ back to $x_0$, closing up the loop.

A: This answer is a complement to Andy's one. If $X$ and $Y$ are complex algebraic varieties then you have the following fact (see more generally Kollár "Shafarevich maps and automorphic forms" Proposition 2.10.2):
If $X$, $Y$ are irreducible algebraic varieties with $Y$ normal and $f:X\to Y$ is a dominant morphism such that the geometric generic fiber is connected then $f_*:\pi_1(X)\to \pi_1(Y)$ is surjective.
In case $Y$ is not normal then the above fact is not true. Take $Y=$ nodal cubic, $X=$ normalization of $Y$ minus one of the two preimages of the node. This situation is realized topologically as follows: it's the map from the sphere minus North Pole to the sphere with North Pole and South Pole identified. (Deleting one of the two points is not necessary to provide a counterexample to the above fact but it provides a counterexample with connected fibers, which is the case you are interested in).
