Let $X$ be a topological space and $\dfrac{X^n}{S_n}$ symmetric product space generated by the action of the symmetric group of $n$ letters $S_n$ (with $n\geq2$) on $X^n$. It's clear that if $X$ is path-connected (resp. arc-connected), then $\dfrac{X^n}{S_n}$ is also path-connected (resp. arc-connected). My question is about the converse, maybe someone have a hint to prove that's true or a counterexample for show that's false.

  • $\begingroup$ Well, for nice (locally path-connected) spaces it is true and easy since path-connectedness is equivalent to connectedness, and it is clear that you can turn a clopen subset of $X$ into a clopen subset of the symmetric power. $\endgroup$ Dec 28, 2019 at 3:39

1 Answer 1


Here is a counterexample. Let $C=\{(x,\sin(1/x)):x\in(0,1]\}\cup\{0\}\times[-1,1]$ be the topologist's sine curve and let $X$ be the quotient space $C/{\sim}$ where $\sim$ is the equivalence relation that collapses $\{0\}\times(-1,1)$ to a point. Then $X$ is not path-connected, by the essentially same argument as for $S$. However, I claim $X^2/S_2$ is path-connected.

To prove this, let $z\in X^2/S_2$ be the point $[(0,-1),(0,1)]$. Consider a map $f:[0,1)\to X^2/S_2$ which has the two points alternating moving left along the sine curve, such that at any time, one of the points has second coordinate $0$. I claim that $f(t)\to z$ as $t\to 1$. Indeed, a neighborhood of $z$ consists of all pairs of points such that one is in some given neighborhood $U$ of $\{0\}\times [-1,1)$ and the other is in some given neighborhood $V$ of $\{0\}\times(-1,1]$. Both $U$ and $V$ will contain all points $(x,0)$ for $x$ sufficiently small, and thus for $t$ sufficiently large, one of the points of $f(t)$ is in $U\cap V$. But $U\cup V$ is a neighborhood of $\{0\}\times [-1,1]$ and thus contains an entire rectangle $[0,\epsilon]\times[-1,1]$ by compactness, and so both points of $f(t)$ are in $U\cup V$ for sufficiently large $t$. It follows that for all sufficiently large $t$ we can assign one of the points of $f(t)$ to $U$ and the other to $V$ (since one is in $U\cap V$ and the other is in $U\cup V$). That is, $f(t)$ is eventually in every neighborhood of $z$.

Thus, $f$ can be extended to give a path from pairs of points in the sine curve to $z$. Similarly, you can get a path from a pair of points where one is in the sine curve and one is in $\{0\}\times[-1,1]$ to $z$ (just let the point in $\{0\}\times[-1,1]$ always be in $\{0\}\times (-1,1)$ so it is always in $U\cap V$ as in the argument above). It follows easily that every point of $X^2/S_2$ is connected to $z$ by a path, so $X^2/S_2$ is path-connected.

On the other hand, it is true if $X$ is Hausdorff; let me first sketch a proof for $n=2$. Suppose $f:[0,1]\to X^2/S_2$ is a path; I claim we can lift $f$ to a path in $X^2$. Let $\Delta\subset X^2$ be the diagonal, which is closed since $X$ is Hausdorff, and let $p:X^2\to X^2/S_2$ be the quotient map. It is easy to check that $p$ restricts to a homeomorphism $\Delta\to p(\Delta)$, so we can first lift $f$ on the set $f^{-1}(\Delta)$. Now $f^{-1}(\Delta)$ is closed in $[0,1]$ so its complement is a disjoint union of open intervals. Moreover, $p$ restricts to a covering map $X^2\setminus \Delta\to p(X^2\setminus \Delta)$ (this is easy to prove using the fact that $X$ is Hausdorff; any open set defined by a pair of disjoint open sets around the two coordinates will be evenly covered). So, on each interval of $[0,1]\setminus f^{-1}(\Delta)$ we can lift $f$ by the lifting theorem for covering spaces. Combining these with our lift on $f^{-1}(\Delta)$, we have a lift of $f$ on all of $[0,1]$, but we still need to check that it is continuous as we approach a point $t\in f^{-1}(\Delta)$ from $[0,1]\setminus f^{-1}(\Delta)$. This is easy though because both points of $f(s)$ must be approaching the single point of $f(t)$ as $s\to t$.

For $n>2$, it is similarly true that any path in $X^n/S_n$ can be lifted to $X^n$ for any $n$ if $X$ is Hausdorff. The proof is similar to the case $n=2$ but the details are much more intricate. Instead of just splitting into the diagonal and its complement, you stratify $X^n/S_n$ by the sets $\Delta_k$ which consist of of points with exactly $k$ distinct coordinates. Using the fact that $X$ is Hausdorff, you can show $p$ is a covering map over each $\Delta_k$. Then you can lift $f$ on the sets $f^{-1}(\Delta_k)$ one by one, but you must do so very carefully to make sure they combine to be continuous. (The last step of the argument for $n=2$ where you check continuity is no longer automatic, and you have to choose your new lift and possibly also adjust your old ones to make them combine to be continuous, using a compactness argument to do so in finitely many steps.)

  • $\begingroup$ Thanks for the answer, I understand the general idea of your counterexample and the proof for the Hausdorff case. But I have two question about the counterexample. The first is for the explicit definition of $f$ an the second is if the same counterexample yields for the arcwise connected case. $\endgroup$
    – Ludwik
    Dec 28, 2019 at 17:25
  • $\begingroup$ Writing down $f$ explicitly is a bit messy but here are more details. Let $x_0,x_1,x_2,\dots$ be the points of the sine curve with second coordinate $0$, in decreasing order of first coordinate. The path $f(0)$ then starts at $[x_0,x_0]$, then moves to $[x_0,x_1]$ along the sine curve, then to $[x_1,x_1]$, then to $[x_1,x_2]$, then to $[x_2,x_2]$, and so on. In each step, only one of the two points moves, while the other one stays constant at $x_n$. $\endgroup$ Dec 28, 2019 at 17:30
  • $\begingroup$ I don't think $X^2/S_2$ is arc-connected because there won't be an arc between $[(0,1),(0,1)]$ and $[(0,-1),(0,-1)]$. You could make it arc-connected though by just adjoining to $X$ paths between each pair of points in $\{0\}\times[-1,1]$. Or more simply, just add the entire rectangle $[-1,0]\times [-1,1]$ to $C$ (and still collapse just $\{0\}\times (-1,1)$ to a point to get $X$). $\endgroup$ Dec 28, 2019 at 17:34
  • $\begingroup$ Thanks. Now I understand the behavior of $f$ and I can try make the counterexample for the arcwise connected case. $\endgroup$
    – Ludwik
    Dec 28, 2019 at 18:46
  • $\begingroup$ Casually I found your answer in this post and I think about metrizability result for my finite case. You know some result about it? Can give me some references for further reading? mathoverflow.net/questions/193988/… $\endgroup$
    – Ludwik
    Dec 28, 2019 at 22:53

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