Homotopy equivalence between circles I'm wondering about one thing: let's consider a plane with one hole $ \mathbb{R}^2 \setminus \{0\} $. I'm wondering whether the two subsets:
$$ S^1 = \{(x,y) \in \mathbb{R}^2 \setminus \{0\}: x^2 + y^2 = 1\} $$
$$ S^1 + (0,2) = \{(x,y)\in\mathbb{R}^2\setminus\{0\}: x^2 + (y-2)^2 = 1\}$$
Are homotopy equivalent. 
I'm actually wondering about the definition, because they are supposed not to be homotopy equivalent, I just can't get that from the definition: the map $ f : S_1 \rightarrow S_1 + (0,2), ~ f(x,y) = (x, y + 2) $ is continuous, open, bijective.
What am I getting wrong?

The two sets viewed as topological spaces (with the subspace topology) are homeomorphic and therefore in particular homotopy equivalent.
But they are not homotopic when we view them as subsets of the topological space $\mathbb R^2\setminus \{0\}$.
"Homotopy equivalence" and "homotopy" are two different equivalence relations -- one applies to two topological spaces in general, the other is between two subsets of one topological space (or strictly speaking between two maps whose images the two subsets are).
