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?]

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    $\begingroup$ As it may puzzle some readers: "topological fundamental group" should be understood as "fundamental group" in the sense of topology. It's used in complex geometry to distinguish with the étale fundamental group. It's unrelated to the topological fundamental group in any meaning where it refers to some topology on the fundamental group (which is relevant for spaces that are locally complicated such as Hawaiian earrings). $\endgroup$
    – YCor
    Jun 8, 2020 at 11:10
  • $\begingroup$ @YCor Yes, thanks for clarification! (Here is a related question on etale fundamental group mathoverflow.net/questions/223885/… where the answer is positive) $\endgroup$
    – user39380
    Jun 8, 2020 at 11:29
  • $\begingroup$ In 7,2,9 of my book "Topology and Groupoids" (pdf availalble) there is an exact sequence of a fibration of groupoids which also deals with some extra information on $\pi_0$ and actions which it would be interesting if you found relevant.. $\endgroup$ Jun 8, 2020 at 15:50

2 Answers 2


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).

  • $\begingroup$ Thanks for the reference! $\endgroup$
    – user39380
    Jun 9, 2020 at 1:19

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:

  1. Start at $x_0$.
  2. 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 $E_1$.
  3. 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 $E_2$.
  4. etc.
  5. 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.
  • $\begingroup$ There is some typo where $X$ and $Y$ are flipped? Thanks for the explanation! $\endgroup$
    – user39380
    Jun 9, 2020 at 1:43
  • $\begingroup$ As pointed out in the answer, by reduction to 1-skeleton, one do not really need to lift arbitrary paths. Is there a general condition (properness seems too strong for quasi-proj vars) that there exists a CW-decomposition of algebraic varieties, so that all 1-simplices can be lift? (For example, in the normalization of nodal curve, this does not hold, I am not sure to what generality this method can be applied?) $\endgroup$
    – user39380
    Jun 9, 2020 at 1:56
  • $\begingroup$ @Qixiao: I fixed the typos, sorry about them! You can definitely make this work if the varieties in question are projective, but I don't know an easily quotable result in the noncompact case. $\endgroup$ Jun 9, 2020 at 15:38

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