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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.

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    $\begingroup$ I think you want "path-connected" everywhere you say "connected" or there are some obnoxious counterexamples, e.g. involving the Warsaw circle. $\endgroup$ – Qiaochu Yuan Oct 10 '18 at 4:23
  • $\begingroup$ this is a great question! $\endgroup$ – Andres Mejia Oct 10 '18 at 4:27
  • $\begingroup$ @QiaochuYuan yes, thank you $\endgroup$ – Dmitry K Oct 10 '18 at 4:27
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    $\begingroup$ Note also that the desired statement is true if $f : X \to Y$ is a fibration (e.g. a fiber bundle), by the long exact sequence in homotopy: en.wikipedia.org/wiki/… . $\endgroup$ – Qiaochu Yuan Oct 10 '18 at 4:27
  • $\begingroup$ And fibrations are very common; for example by Ehresmann's fibration theorem (en.wikipedia.org/wiki/Ehresmann%27s_lemma) every surjective submersion between closed smooth manifolds is a fibration. So counterexamples would need to look a bit strange. $\endgroup$ – Qiaochu Yuan Oct 10 '18 at 4:35
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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.

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  • $\begingroup$ Nice! Perhaps the original poster would like to add that everything is locally path-connected, to rule out this example. :-) $\endgroup$ – Hew Wolff Oct 10 '18 at 15:10
  • $\begingroup$ for $\pi_1$, that just means locally path connected, which is a pretty light assumption I guess. $\endgroup$ – Andres Mejia Oct 10 '18 at 15:57
  • $\begingroup$ @HewWolff the spaces are supposed to be CW complexes anyway $\endgroup$ – Dmitry K Oct 10 '18 at 16:06
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    $\begingroup$ @DmitryK Keep in mind this doesn't allow arbitrary continuous maps (or even cellular) between CW complexes, as you make local path connectedness assumptions on the fiber above every point. (doesn't bother me, but worth noting.) $\endgroup$ – user98602 Oct 14 '18 at 3:59
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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.

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  • $\begingroup$ @DmitryK, I know next to nothing about algebraic geometry, but it seems likely that the PL category is good enough for you. For example it appears from the paper Triangulation of Locally Semi-Algebraic Spaces that algebraic varieties are triangulable. $\endgroup$ – Hew Wolff Oct 10 '18 at 18:34
  • $\begingroup$ Algebraic maps are certainly not PL, though. One would like to make a similar argument in that situation, since surjectivity of a map on $\pi_1$ is a homotopy invariant property, you may homotope to a PL approximation of your algebraic map. But can one do this in such a way that the new map still has connected fibers? $\endgroup$ – user98602 Oct 11 '18 at 17:22
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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.

enter image description here

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.

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  • $\begingroup$ $X$ is not compact and fibers are discrete $\endgroup$ – Dmitry K Oct 10 '18 at 3:51
  • $\begingroup$ $\mathbb{R}$ is not compact. $\endgroup$ – qualcuno Oct 10 '18 at 3:51
  • $\begingroup$ Oh I didn't see that. $\endgroup$ – J1U Oct 10 '18 at 3:54
  • $\begingroup$ This is the same as one of the deleted answers and has the same problem: the fiber over exactly one point fails to be connected (it contains $0$ and $1$). $\endgroup$ – Qiaochu Yuan Oct 10 '18 at 4:13
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    $\begingroup$ Hi J1U, good idea but actually $f^*(1, 0) = 1$, not $2$. For example, take $C$ to be the subspace of $T^2$ given by the bottom edge of your square diagram: $f$ maps it homeomorphically to $S^1$, so $f^*$ is onto. More generally, your construction is too "nice": it is a fiber bundle, so following @QiaochuYuan's comments above, we will need something weirder! $\endgroup$ – Hew Wolff Oct 10 '18 at 15:46

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