Topological quotients of $\mathbb{R}^2$ by subspaces

Question

Let $X$ be a topological space and $A \subseteq X$. Then, let $X_A$ denote $X/R$, where:

$$R = \{ (x,y) : x = y \text{ or } (x,y) \in A \times A\}$$

i.e. we collapse $A$ to a point. I will call $X_A$ the quotient of $X$ by $A$.

Conjecture: Let $X = \mathbb{R}^n$ and $K \subseteq X$ be compact and path-connected with path-connected complement. Then, $X_K$ is homeomorphic to $X$.

You may note that, for the case $n=2$, this is similar to the Jordan curve theorem, which I don't want to assume.

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I think that this is true, but it's quite difficult to construct the required homeomorphism; I think I would need to brush up my analysis first. Equally, if it's not true, I don't have the imagination to guess what the counterexample looks like. I've only proved a basic lemma:

Lemma: Let $Y,Z$ be topological spaces. Let $U_n \subseteq Y$ be an increasing family of open sets such that $\bigcup_n U_n = Y$ and for each $n$, there exists $f_n$ such that

• $f_n : U_n \to Z$ is a homeomorphism onto its image
• $f_n \restriction_{U_k}=f_k$ for any $k \le n$
• $\bigcup_n f_n(U_n) = Z$

Then, $Y \cong Z$.

Of course, this is nothing astonishing; the proof is just to take the union of all the homeomorphisms, so to speak. But it should simplify the proof of the conjecture.

Answer 1

Here is an answer clarifying and correcting my earlier comment.

1. For each $n\ge 3$, there exists a subset $A\subset R^n$ homeomorphic to $[0,1]$ (hence, path-connected with connected complement) such that $Q=R^n/A$ is not a topological manifold, hence, is not homeomorphic to $R^n$. The example is essentially due to Bing: Take a (wild) arc $A$ embedded in $R^n$ such that $R^n\setminus A$ is not simply-connected. Then $Q$ is simply-connected but removing a point from $Q$ (namely, the projection of $A$) makes it non-simply-connected. Hence, $Q$ cannot be a topological manifold.

2. Suppose that $A\subset R^2$ is a compact such that its complement is simply-connected (which follows from your assumptions but is strictly weaker). Then $R^2/A$ is homeomorphic to $R^3$.

This is an application of several nontrivial results which will also require several definitions.

(a) A compact metrizable space is said to be cell-like if for some (equivalently, every) embedding $i$ of $A$ in the Hilbert cube $H$ has the property that for every neighborhood $U$ of $i(A)$ in $H$, the inclusion $i(A)\to U$ is null-homotopic. For instance, a closed ball is cell-like but the circle is not.

This definition is a bit hard to digest. But there is a sufficient condition for $A$ to be cell-like:

(b) A compact subset $A\subset R^k$ is said to be cellular if there is a decreasing sequence of neighborhoods $U_j$ of $A$ in $R^k$ such that:

(i) Each $U_j$ is homeomorphic to $R^k$.

(ii) $\bigcap_{j} U_j=A$.

As it turns out, if $A\subset R^k$ is compact and $R^k\setminus A$ is homeomorphic to $R^k$, then $A$ is cellular. This theorem is due to M.Brown:

Brown, Morton, The monotone union of open $n$-cells is an open $n$-cell, Proc. Am. Math. Soc. 12, 812-814 (1961). ZBL0103.39305.

In particular, if $A\subset R^2$ is a compact with simply-connected complement, then $A$ is cellular.

Lastly, R.Moore proved in

Moore, R. L., Concerning upper semi-continuous collections of continua., Trans. Am. Math. Soc. 27, 416-428 (1925). ZBL51.0464.03.

that if $A\subset R^2$ is cell-like then $R^2/A$ is homeomorphic to $R^2$ (in fact, he proves much more but you do not need this).

See also

Edwards, Robert D., The topology of manifolds and cell-like maps, Proc. int. Congr. Math., Helsinki 1978, Vol. 1, 111-127 (1980). ZBL0428.57004.

for more details and references.