TITLE: Surjectivity of a continuous map implies surjectivity on $\pi_1$
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.