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?
 A: 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).
A: From Makholm's definition, a homotopy between two subspaces is a continuous deformation from one such subspace to another (i.e. an open set, the trace of a graph, the image of a function, etc...). If we use the time interval $t = [0,1]$ for the purpose of parametrizing the deformation, then at $t=0$ the deformation 'equals' the first subspace, and at $t=1$ the deformation 'equals' the second. The fact that the homotopy is continuous is important, as highlighted by your example. If there exists a homotopy between two subspaces, they are said to be homotopic.
Using @Makholm's point above, that you in fact mean 'homotopic', you make an incorrect assertion that "$f : S_1 \rightarrow S_1 + (0,2), f(x,y) = (x, y + 2)$ is continuous, open, and bijective" implies that the two subspaces are homotopic. To define it explicitly, let the homotopy $F(x,t)$ be defined such that $F(x,0) = x \in S^1$, $F(x,1) = f(x) \in S^1 + (0,2)$ for $t \in [0,1]$. The straight-line homotopy should suffice as a counterexample, as the punctured plane is still path connected. But if $F(x,t) = (1-t)x + t\cdot f(x)$, if we take $x = (0,-1) \in S^1$, we cannot move $x$ along the $y$-axis through $0$ without introducing a discontinuity. Proving that there is no such homotopy between your subspaces, however, is a little more difficult, and requires additional machinery to prove.
To stay on topic, homotopy equivalence would be trivial here, because if we treat these two entities as spaces with the topologies they inherit from $\mathbb{R}^2$, then these two spaces are homeomorphic. So we take $f: S^1 \to S^1 + (0,2)$ to be translation by $2$ units, and $g = f^{-1}$, then  $g \circ f = id_{S^1}$ and $f \circ g = id_{S^1 + (0,2)}$. In fact, there is a stronger result here using the same approach. ANY two homeomorphic spaces are homotopy equivalent, since we can take the original homeomorphism $f$, and compose it with its (necessarily) continuous inverse $g ~ (=f^{-1})$ to yield the identities on their respective domains.
EDIT: I see in the time I've typed this, @Makholm has elaborated on some of his points. I won't remove this post because it provides a little more rigor to your case, though I did try and simplify for you the definition of 'homotopic' relative to the wikipedia entry up front. 
