gn.general topology – Isotopy Classes of Non-Connected Planar Sets

I was just looking through some of my old questions on StackExchange and noticed that this one went totally unresolved. I still think it’d be useful as a lemma for plane topology if it were true, and would be curious to see a counterexample if not.

Suppose we are given two pairs $(U, V)$ and $(Y, Z)$ of connected, simply connected, bounded, open sets in the plane, where $U cap V = Y cap Z = varnothing$. However, their boundaries may intersect. Assume their closures are also simply connected. Let $A = U cup V$ and $B = Y cup Z$. Suppose we also know the following, where ‘$simeq$‘ refers to planar isotopy:

  1. $U simeq Y$

  2. $V simeq Z$

  3. $bar{U} cap bar{V} simeq bar{Y} cap bar{Z}$

Then this is not enough to imply that $A simeq B$, due to an example at the bottom of this post.

So I’m wondering, what are some of the simplest conditions we can add on to those three above to ensure that $A simeq B$?

  1. Is it enough that $bar{A}$ is homeomorphic to $bar{B}$?

  2. Alternatively, assume $bar{U} setminus bar{V}$ is homeomorphic to $bar{Y} setminus bar{Z}$. Would assuming this, as well as its obverse for $bar{V} setminus bar{U}$ and $bar{Z} setminus bar{Y}$, be sufficient?

Counterexample, as promised: Let $U = Y$ be a ‘thickened open topologist’s sine curve’, i.e. a shrinking tubular neighborhood of the standard topologist’s sine curve with limit arc being $(-1, 1)$ on the $y$-axis (though not itself containing this arc, just the ‘wiggly part’ on $(0, pi)$). Let $V$ be the reflection of $U$ across the $y$-axis, and let $Z$ be the same as $V$ except scaled by $frac{1}{2}$ vertically. So in other words $Z$ is just a vertically-squished topologist’s sine curve coming from the other direction. Then clearly $U, V, Y, Z$ satisfy all of the above conditions (condition 3 is satisfied since each is just an arc, albeit of different lengths) but $A notsimeq B$.

Obviously, this problem could be generalized in all sorts of ways, though Lakes of Wada phenomena would be a complication if you start increasing the number of components in $A$ and $B$.