Annulus without ambient space?

Question

An annulus $A = \{(x,y) \in \mathbb{R}^2: 1<x^2+y^2 < 2\}$ is a topological subspace of 2-D Euclidean space $\mathbb{R}^2$ and has a boundary composed of two components. But if we forget $\mathbb{R}^2 - A$ completely (so the whole universe is $A$ itself), then it clearly has no boundary. This space should be topologically different from $A$. However, in terms of loops (I just started studying elementary algebraic topology), the two spaces seems to be equivalent. Intuitively the two spaces have the same classes of continuously deformable loops. With what algebraic topological tool can I distinguish the two spaces?

Answer 1

First of all, "loops" will not detect something as subtle as you mean. However, it is also true that taking the closure of an open subset and considering the boundary of the resulting space depends on how you embed a space.

For example the subsets $\{(x,y,z) \mid z \neq \pm 1\} \subset S^2$ is topologically a cylinder $S^1 \times \mathbb R$ which is homeomorphic to the open annulus. Likewise, you can consider $\{(x,y,z) \mid |z|<1/2\}$ is also an annulus. The "boundary" of the first is two points and the boundary of the latter is $S^1 \coprod S^1$, despite them being homeomorphic.

On the other hand "loops" will only detect up to homotopy. For example, as a space unto itself, $S^1$ and $S^1 \times \mathbb R$ will have the same fundamental group.