Let ($X$, $x_0$) be a topological space with a base point, and denote the fundamental group of $X$ as $\pi_1(X)$. Let $N$ be a normal subgroup of $\pi_1(X)$.

Does there necessarily exist an equivalence relation $\thicksim$ on $X$ such that $\pi_1(X/\thicksim)$ is isomorphic to $\pi_1(X)/N$?

  • $\begingroup$ Duplicate of math.stackexchange.com/questions/2036085/… $\endgroup$
    – Igor Rivin
    Feb 17, 2018 at 4:13
  • 8
    $\begingroup$ @IgorRivin This is not the same question at all...! $\endgroup$ Feb 17, 2018 at 17:22
  • $\begingroup$ @NajibIdrissi Do tell! $\endgroup$
    – Igor Rivin
    Feb 17, 2018 at 22:49
  • 3
    $\begingroup$ This is kind of stupid, but here it is. You can use space-filling curves to construct a surjective map from $[0,1]$ to any connected, finite CW-complex $L$. If $f: K \to L$ is any map of finite CW-complexes, $L$ is connected, and $K$ is not $0$-dimensional, you can use this find a homotopy from $f$ to another map $g$ that does a space-filling curve on a small copy of an interval inside $K$, making $g$ surjective. This a continuous surjective map of compact Hausdorff spaces, so you've make $L$ homeomorphic to a quotient space of $K$, realizing the map $\pi_1(K) \to \pi_1(L)$ you started with... $\endgroup$ Feb 18, 2018 at 0:03

2 Answers 2


This is not always possible. (This answer was worked out in collaboration with Raymond Cheng.)

Example. Let $X$ be the pseudocircle: it is the finite quotient of the unit circle $S^1 \subseteq \mathbb R^2$ where the open upper half segment is contracted to a point and the lower half segment is contracted to a point:


It is a four-point topological space with two closed points and two open points, and it is weakly homotopy equivalent to $S^1$. In particular, $\pi_1(X) = \mathbb Z$.

On the other hand, there are only finitely many possible quotients $X/{\sim}$, and therefore not every group $\mathbb Z/n\mathbb Z$ can occur as the fundamental group of $X/{\sim}$. $\square$

Remark. We can actually go through all possible quotients. If $\sim$ is a nontrivial equivalence relation, then $X/{\sim}$ is connected and has fewer than $4$ points. In fact, it is always contractible:

  • Any one-point space $Y$ is contractible.
  • Any connected two-point space $Y$ is either Sierpiński space or indiscrete, both of which are contractible¹.
  • If $|X/{\sim}| = 3$, then we have either:
    • identified two open points to give the 'pseudo-closed-interval', which is contractible¹;
    • identified two closed points to give a 'pseudo-open-interval', which is contractible¹;
    • identified a closed point with an open point, giving a three-point space where one point is open, one point is closed, and one point is neither. This is the finite $T_0$ space corresponding to a total order, and it is contractible¹.

¹For example, consider the 'pseudo-closed-interval' $Y$:


Labelling the points on $Y$ by $0, \varepsilon$, and $1$, we can contract $Y$ to a point by \begin{align*} \Phi \colon Y \times [0,1] &\to Y \\ (y,t) &\mapsto \left\{\begin{array}{ll} y, & t = 0 \\ 0, & y = 0, \\ \varepsilon, & 0 < t < 1 \text{ and } y \neq 0, \\ 0, & t = 1. \end{array}\right. \end{align*} The inverse image of $0$ is the closed set $(Y \times \{1\}) \cup (\{0\} \times [0,1])$, and the inverse image of $1$ is the closed set $\{(1,0)\}$, proving that $\Phi$ is continuous. Hence, $\Phi$ is a homotopy from $\operatorname{id}_Y$ to the constant map $0$, showing that $Y$ is contractible. The proofs for the other spaces are similar.

  • 1
    $\begingroup$ It might help in the checking of the proof that the fundamental group depends only on the weak homotopy type and that every finite topological space is weakly equivalent to the nerve of its specialization poset. $\endgroup$ Mar 24, 2018 at 9:54
  • $\begingroup$ @DenisNardin: of course for the question at hand, one can also easily compute the fundamental group of $X$ by hand, in exactly the same way that one does for $S^1$. $\endgroup$ Mar 24, 2018 at 15:37

Not exactly an answer to your question, but it might be interesting anyway. Suppose your space $X$ is nice enough (path connected, locally path connected, semi-locally simply connected). Then under the Galois correspondence, the normal subgroup $N \triangleleft \pi_1(X)$ corresponds to a path-connected cover $p : Y \to X$ such that $\pi_1(Y) = N$ and $p_* : \pi_1(Y) \to \pi_1(X)$ is the inclusion.

Consider the mapping cylinder $M_p = (Y \times [0,1]) \cup_p X$. The covering $p : Y \to X$ factors as $Y \xrightarrow{\tilde{p}} M_p \xrightarrow{r} X$, where $r$ is a homotopy equivalence (a deformation retract). Therefore you have $\pi_1(M_p) = \pi_1(X)$ and $\tilde{p}_*$ induces the same map as $p_*$, i.e. the inclusion.

Finally, you have the long exact sequence of relative homotopy groups: $$\dots \to \pi_1(Y) \to \pi_1(M_p) \to \pi_1(M_p, Y) \to \pi_0(Y) \to \pi_0(X) \to \dots$$ and since $X$ and $Y$ are connected, $\pi_0(Y) \to \pi_0(X)$ is an isomorphism. Thus $$\pi_1(M_p, Y) = \pi_1(M_p) / \pi_1(Y) = \pi_1(X)/N.$$ Finally if your spaces are super-nice (CW-complexes), then there is a surjection $\pi_1(M_p, Y) \to \pi_1(M_p / Y)$, where $M_p / Y$ is the mapping cone of $p$. So you can view $\pi_1(X)/Y$ as a quotient of the fundamental group of a quotient of a space which deformation retracts onto $X$. I don't know (think) that you can say something better than that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.