Is there a relationship between a quotient group of the fundamental group of X and the fundamental group of a quotient topology of X?

Question

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 $\pi_1(X)/N$?

Answer 1

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 $X/{\sim}$. $\square$

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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.

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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.

Answer 2

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.