# Removing homeomorphic connected components

connectednessgeneral-topology

Suppose that topological spaces $$X$$ and $$Y$$ are homeomorphic, and let $$C_1 \subset X$$ and $$C_2 \subset Y$$ be connected components in each of the respective spaces, so that $$C_1$$ is homeomorphic to $$C_2$$ with the subspace topologies. Is it true that $$X \setminus C_1$$ is homeomorphic to $$Y \setminus C_2$$ ?

My thoughts: if $$f: X \mapsto Y$$ is an homeomorphism, then it maps connected components to connected components. We have two possibilities: if $$f(C_1) = C_2$$ then $$f|_{X \setminus C_1}$$ is the desired homeomorphism. Otherwise, let $$C_3 \subset X$$ and $$C_4 \subset Y$$ be the components such that $$f(C_1) = C_4$$, $$f(C_3) = C_2$$ and $$C_1 \neq C_3$$, $$C_2 \neq C_4$$. Define $$h: X \setminus C_1 \mapsto Y \setminus C_2$$ as:

$$h(x) = \begin{cases} x & x \notin C_3 \\ f(g(f(x))) & x \in C_3 \end{cases},$$

where $$g : C_2 \mapsto C_1$$ is an homeomorphism. The resulting map $$h$$ is well defined and bijective, maps connected components to connected components and restricts to an homeomorphism in each connected component. Does this suffice to conclude that $$h$$ is an homeomorphism?

With respect to this latter question, I think that you can take an open set $$U \subset Y \setminus C_2$$, which is just any open subset in $$Y$$ since $$Y \setminus C_2$$ is open, and one has $$h^{-1}(U) = \bigcup_{j}h^{-1}(U) \cap X_j = \bigcup_{j}h^{-1}(U \cap Y_j)$$, where $$X_j$$ are the connected components of $$X \setminus C_1$$ and $$Y_j = h(X_j)$$ those of $$Y \setminus C_2$$, but every $$h^{-1}(U \cap Y_j)$$ is open in $$X_j$$ since $$U \cap Y_j$$ is open in $$Y_j$$ (since $$h$$ restricted to an homeomorphism on each component), and being open in a connected component amounts to being open in the whole space, so $$h^{-1}(U)$$ is a union of open sets in $$X \setminus C_1$$ and so it's open.

The proposition (which ammounted to showing that a bijective mapping which restricts to a homeomorphism in each connected component is a homeomorphism as a whole) turns out to be true if the connected components of the original space, $$X$$, are open. This is to ensure the correctness of: "being open in a connected component ammounts to being open in the whole space", which I used in my original line of reasoning.
As for a counterexample when connected components are not open, Rob Arthan gave the following one: consider $$X = Y = \mathbb S^1 \sqcup (\mathbb Q \times \mathbb S^1)$$, $$C_1 = \mathbb S^1$$ and $$C_2 = \{0\} \times \mathbb S^1$$. Obviously $$C_1 \simeq C_2$$, but $$X \setminus C_1 = \mathbb Q \times \mathbb S^1$$, while $$Y \setminus C_2 = \mathbb S^1 \sqcup (\mathbb Q \setminus \{0\} \times \mathbb S^1)$$, which cannot be homeomorphic because $$\mathbb S^1 \subset Y \setminus C_2$$ is a connected, open subset, but there are no such sets in $$X \setminus C_1$$.