How to call two subsets that can be deformed into each other? Given a topological space $X$, is there a canonical name for the equivalence relation generated by the following relation on the subsets of $X$?
$A \sim B :\Leftrightarrow \exists \text{ continuous } h:[0,1]\times X \to X,\; h(0, •) = id_X,\; h(1, A) = B$
that is $A$ can be “homotopically deformed” to match $B$.
 A: This very closely matches the notion of homotopy.  However, homotopy is slightly different, as it represents the notion of a map $h\colon [0,1] \times A \rightarrow X$ such that $h(0,-)$ is the identity and $h(1,-) = B$.  This is basically a deformation of maps into $X$, but doesn't require that all of $X$ be mapped at each step.  Technically, you wouldn't say the sets $A$ and $B$ are homotopic, because being homotopic is a property of maps, not sets, so you would say the inclusion of $A$ into $X$ is homotopic to a surjective map $A \rightarrow B$.  But many people use the term "homotopic sets" as shorthand for this.
For many familiar spaces, being homotopic is equivalent to the condition you define.  However, the notions are not equivalent for all spaces - for example, consider the union of the negative real numbers and the positive rationals.  Then the singleton sets $\{0\}$ and $\{-1\}$ are homotopic, but do not satisfy your property.
There's also the notion of ambient isotopy, which does specify a map $h\colon [0,1] \times X \rightarrow X$ that covers the full space, but two sets being ambient isotopic is stronger than what you've stated, as it requires the maps $h(t, X)$ to be homeomorphisms for each fixed $t$.  I don't know of a term that only requires them to be continuous.
A: I would add to the previous answer that f $A \subset B$ or $B \subset A$ and the application $h|_A=Id_X$ (assuming that $A \subset B$) then you have what is called a $\textbf{Deformation retract}$ (http://en.wikipedia.org/wiki/Deformation_retract) and the application is called a retraction.
This has some interesting properties. The most important one is that, if $A$ is a deformation retract of $B$ then both $A$ and $B$ have the same fundamental group. A common example of this is to retract the cylinder to a circunference along their generatrices. So, the circunference and the cylinder have the same fundamental group:
$$ \pi(\mathbb{S}^1)=\pi(Cylinder)=\mathbb{Z} $$
P.D. I would have add this as a comment rather than an answer but I can't post comments yet.
