Surjectivity of a continuous map implies surjectivity on $\pi_1$ Let $X\to Y$ be a continuous surjective map between path-connected compact topological spaces (say, CW complexes), such that every fiber is path-connected. Can it be true that it always induces the surjective homomorphism between $\pi_1(X)$ and $\pi_1(Y)$?
I was unable to find a counterexample.
The only way to prove it that comes to my mind is to use cellular approximation, but it can destroy connectedness of fibers when we deform the map.
 A: No for general spaces.
Following Qiaochu Yuan: 
Let $X$ be the Warsaw circle (the version that is path-connected, like this). Let $Y=S^1$. Radially project $X$ to $Y$. The projection cannot induce a surjective homomorphism between the fundamental groups, as $X$ is simply connected. The fibers of the projection maps over every point is a singleton except for one exceptional point where the fiber is a line. 
Edit:
The main theorem of this paper of Smale seems to answer your question. (I think $LC^n$) means that it is locally $n$-connected, but Smale doesn't seem to define this.  
A: One good way to rule out pathologies such as the Warsaw circle is to restrict ourselves to the piecewise linear category. I think that in this case, the map is always onto in $\pi_1$.
Suppose that $\gamma: S^1 \to Y$ is a loop in $Y$. Since everything is PL, look at the original map $f$ as a map from the $1$-skeleton of $X$ to the $1$-skeleton of $Y$, a homomorphism of undirected graphs. $\gamma$ is now a finite loop of of directed edges in $Y$. Because $f$ is onto, each of these edges can be lifted to $X$. Say we have two consecutive edges $y$ and $y'$, where $y = (y_0, y_1)$, $y' = (y'_0, y'_1)$, and $y_1 = y'_0$. These lift to $(x_0, x_1)$ and $(x'_0, x'_1)$.
But also, the fiber $f^{-1}(y_1)$ is connected, so there is a path in that fiber from $x_1$ to $x'_0$. So we can stitch together the lifted edges to get a finite loop $\beta$ in $X$ which is a lift of $\gamma$ up to homotopy: $f \circ \beta \sim \gamma$. Now at the level of $\pi_1$, $f^*([\beta]) = [\gamma]$, so $f^*$ is onto.
A: I hope this time it is the right counterexample. Let $X=T^2$, $Y=S^1$. They are both compact, path connected, and even Hausdorff, second countable, locally Euclidean, blah blah. We all know that $T^2$ is a quotient space of $I\times I$ by identifying $(x,0)$ with $(x,1)$ and $(y,0)$ with $(y,1)$. Or since $\mathbb{R}^2$ covers $T^2$, we can consider it as a quotient space of $\mathbb{R}^2$ where $(x,y)$ are identified with $(x+n,y+m)$ for all integers $n$ and $m$. Now consider the image of the line $y=x/2+a$ on $\mathbb{R}^2$ by the map $\mathbb{R}^2 \to T^2$. Then this curve is an embedding of a circle, turning around the torus two times in both directions. Here is the picture for the curve.

Now we construct $f:T^2\to S^1$ by mapping this circle defined by $y=x/2+a$ to a single point in $S^1$. To be specific, let $f(p)=e^{i4\pi a}$ when $p$ lies on the circle $y=x/2+a$. This is well defined since on $\mathbb{R}^2$, every point $(x,y)$ satisfies $y=x/2+a$ for the unique real number $a$. This is obviously surjective. The fiber is path connected―it is just a circle. Also, it is easy to see this map $f$ is continuous.
Let's see how $f^*: \pi_1(T^2) \to \pi_1(S^1)$ looks like. We know that $\pi_1(T^2)=\mathbb{Z}\times \mathbb{Z}$ and $\pi_1(S^1)=\mathbb{Z}$, so it is enough to see what $f^*(1,0)$ and $f^*(0,1)$ is. If you carefully draw the picture, you will see that when you turn a half round around the torus in any direction, the image already take one round in $S^1$. Or more rigorously, you can take lines $x=0$ and $y=0$  on $\mathbb{R}^2$, you can observe how it changes along the map $\mathbb{R}^2 \to T^2 \stackrel{f}{\to} S^1$. They are all two loops around the circle. It means $f^*(0,1)=f^*(1,0)=2$ and the image of $f^*$ is $2\mathbb{Z} \subset \mathbb{Z}$. Hence the induced map $f^*$ is not surjective.

Below are my bad attempts.
Take $X=[0,1]$, $Y=S^1$, and $f:[0,1]\to S^1$ where $f(x)=e^{i2\pi x}$. This is surjective, but $\pi_1([0,1])=0$ and $\pi_1(S^1)=\mathbb{Z}$, and the induced map $f^*:0\to\mathbb{Z}$ cannot be surjective.
Edit: I didn't see $X$ should be compact so I editted the answer.
Edit: It does not also preserve injectiveness. Consider $X=S^1\sqcup S^1$, $Y=T^2=I×I/\sim$ where the relation $~$ is given by $(x,0)\sim(x,1)$ and $(y,0)\sim(y,1)$. Define $f:X\to Y$ by taking one circle in $X$ to ${(x,0)}$ and another circle to ${(y,0)}$. This is an injection. Then the induced map $f^*:F_2\to \mathbb{Z}^2$ (here $F_2$ is a free group of two generator) is the quotient by its commutator, so $f^*(ab)=f^*(ba)$.
Edit: I didn't see that the function should have the continuous fiber, so let me rewrite my answer. Let $S$ be the topologist's sine curve
$$
S=\{(x,\sin(1/x):x\in (0,1]\}
$$
and take the closure of it
$$
\bar S = S\cup \{(0,y):y\in [-1,1]\}
$$
This is connected, compact, but not path-connected. Actually, $X$ is not a CW-complex. Let $X$ be the quotient space of $\bar S$ by identfying $(1,\sin1)$ and $(0,0)$. Now consider a surjection $f: X \to S^1$ defined as following.
$$
f(0,y)=1\ (-1\leq y \leq 1)\\
f(x,\sin(1/x))=e^{2i\pi x}\ (0<x\leq1)
$$ 
This is a continuous map, and the fibers are all connected because it is either a single point or a line segment $\{(0,y)\}$. But $\pi_1(X)=0$ and $\pi_1(S^1)=\mathbb{Z}$, so $f^*$ is not surjective.
I am not sure if the proposition is true when $X$ is CW-complex.
