Homotopy type of the complement of a subspace

I have been trying to prove or disprove the following statement:

Let $A$ and $B$ two subspaces of $\mathbb{R}^2$ with respect to the Euclidean topology. If $A$ and $B$ have the same homotopy type, then $\mathbb{R}^2 \setminus A$ and $\mathbb{R}^2 \setminus B$ have the same homotopy type.

I think the statement is false. Let's consider a subspace $A$ which consists of a single point and a subspace $B$ which consists of a single line. The subspaces $A$ and $B$ have the same homotopy type, because a line is a convex set, so it is contractible. But the subspaces $\mathbb{R}^2 \setminus A$ and $\mathbb{R}^2 \setminus B$ do not have the same homotopy type, because the first one is path connected and the second one is not.

Is this correct?

Solution 1:

It is in fact not true for arbitrary $A,B$. An even simpler counterexample is $A = \{ 0 \}$, $B = \mathbb{R}^2$.

However, if we restrict to plane continua (= compact connected subspaces of $\mathbb{R}^2$), then there is the following complement theorem:

Two plane continua have the same shape if and only if their complements in $\mathbb{R}^2$ are homeomorphic.

This was essentially proved by Karol Borsuk in his paper

"Concerning homotopy properties of compacta." Fundamenta Mathematicae 62.3 (1968): 223-254.

(See also Sher, Richard B. "Complement theorems in shape theory." Shape theory and geometric topology. Springer, Berlin, Heidelberg, 1981. 150-168.)

In this paper Borsuk introduced the concept of shape for compact metric spaces. This is a classification coarser than homotopy type and agreeing with it for "nice" spaces like polyhedra, CW-complexes and ANRs. For example, the closed topologist's sine curve has the shape of a point but is not homotopy equivalent to a point. There are countably many shapes of plane continua which are represented by the finite wedges $W_n$ of $n$ circles (where we understand $W_0$ = one-point space) and the Hawaiian earring $W_\infty$.

If anybody should be interested in the concept of shape, I recommend to have a look into

Mardešic, Sibe, and Jack Segal. Shape theory: the inverse system approach. Vol. 26. Elsevier, 1982.

Note that both conditions (compactness and connectedness) are essential. If we consider closed $A, B \subset \mathbb{R}^2$, then we gave counterexamples in which $A,B$ are connected. Next consider $A = S^1 \cup \{ (0,0) \}$, $B = S^1 \cup \{ (0,2) \}$. These are homeomorphic compact subpolyhedra of $\mathbb{R}^2$. The complement of $A$ consists of two open annuli, the complement of $B$ of an open disk and a punctured open annulus.

Let me finally remark that complements of plane continua are homeomorphic if and only if they have the same homotopy type. This is true because it follows from the above facts that complements of plane continua have finitely many or countably many components, one of them being unbounded and homeomorphic to an open annulus, all other being bounded and homeomorphic to an open disk.