Example of two homeomorphic spaces without a continuous deformation between them? on wikipedia it states:

there need not be a continuous deformation for two spaces to be homeomorphic — only a continuous mapping with a continuous inverse function.

I'm trying to intuitively understand this difference between a continuous mapping and continuous deformation, so I'm trying to find an example of two spaces that are homeomorphic but without a continuous deformation.
Here is a definition of homotopy: 

Formally, a homotopy between two continuous functions f and g from a topological space X to a topological space Y is defined to be a continuous function H : X × [0,1] → Y from the product of the space X with the unit interval [0,1] to Y such that, if x ∈ X then H(x,0) = f(x) and H(x,1) = g(x)

Does the following example satisfy?



In a sense there is no continuous deformation between the two (though I honestly cannot immedately see from the definition why not. i.e. I cannot see why simply letting the different parts of the knot "move through" each other is not allowed by the formal definition of homotopy.)
 A: $\newcommand{\Reals}{\mathbf{R}}$Since this has gotten bumped:
tl; dr: Comparing homeomorphism (a relation between two topological spaces) and homotopy (a relation between two mappings) is a recipe for confusion. Informal, qualitative descriptions of topology do not help, since they frequently use language such as "continuous deformation", which can be interpreted in either sense.

Let $X$ and $Y$ be topological spaces, and let $f$, $g:X \to Y$ be continuous mappings.
Homeomorphism: A mapping $f:X \to Y$ is a homeomorphism if $f$ is continuous and invertible, and its inverse is continuous. If there exists a homeomorphism $f:X \to Y$, we say the topological spaces $X$ and $Y$ are homeomorphic.
Homotopy: In the loosest sense, two continuous mappings $f$ and $g:X \to Y$ are homotopic if there exists a continuous mapping $H:X \times [0, 1] \to Y$ such that $H(x, 0) = f(x)$ and $H(x, 1) = g(x)$ for all $x$ in $X$.
(For technical reasons, it's usually desirable to fix a "basepoint" in each topological space and to assume all mappings send basepoints to basepoints, or in fancy terms to work in the category of pointed spaces. Since the aim here is merely to formalize qualitative descriptions of topology for posterity, we'll ignore basepoints.)
Isotopy: Along the lines of Paul Frost's comment, we can combine these two concepts. First, we say a mapping $f:X \to Y$ is an embedding if $f$ is a homeomorphism onto its image, i.e., an injective mapping with continuous inverse.
If $f$ and $g:X \to Y$ are embeddings, we say $f$ and $g$ are isotopic, or homotopic through embeddings, if there exists a homotopy $H:X \times [0, 1] \to Y$ such that for each $t$ with $0 \leq t \leq 1$, the mapping $H_{t} = H( \cdot, t):X \to Y$ is an embedding.

Let's see how these concepts differ.

*

*The identity mapping $I_{X}:X \to X$ is a homeomorphism. Every mapping $f:X \to Y$ is homotopic to itself. Every embedding $f:X \to Y$ is isotopic to itself. A non-embedding $f:X \to Y$ is not isotopic to anything, because the definition above makes no sense for non-embeddings.


*Any two mappings $f$, $g:X \to \Reals^{n}$ into a Cartesian space are homotopic (because we are working with "unpointed" spaces). The standard argument is to use the straight line homotopy
$$
H(x, t) = (1 - t)f(x) + tg(x) = f(x) + t(g(x) - f(x)),
$$
which connects the points $f(x)$ and $g(x)$ by a line segment, and then simultaneously "deforms" along these segments as $t$ runs from $0$ to $1$.


*Let $X$ denote the unit circle, which we'll represent as the set of points $(x, y) = (\cos\theta, \sin\theta)$ in the Cartesian plane.



*

*The antipodal mapping $g(x, y) = (-x, -y)$ is a homeomorphism of the circle to itself, hence an embedding. The mapping
$$
H((x, y), t) = (x\cos\pi t - y\sin\pi t, x\sin\pi t + y\cos\pi t),
$$
which rotates the circle counterclockwise through angle $\pi t$, is an isotopy from the identity to the antipodal mapping.

*The reflection mapping $g(x, y) = (x, -y)$ is a self-homeomorphism of the circle, but turns out not to be homotopic (much less isotopic) to the identity. To prove this, we need some "homotopy invariant" of mappings, some number or other mathematical structure associated to a continuous mapping in a way that homotopic mappings have the same invariant, and so that these mappings have different invariants. Doing so is beyond the scope of this answer, but a standard choice is winding number. Intuitively, the identity mapping "winds $+1$ times" (i.e., once counterclockwise) while $g$ winds $-1$ times (once clockwise).

*The inclusion mapping of the circle into the plane $\Reals^{2}$ and the reflection mapping of the circle into the plane are homotopic as mappings into the plane. Indeed, Point 2. above guarantees that any two mappings $f$, $g:X \to \Reals^{2}$ are homotopic. One point is, whether or not two mappings are homotopic depends on the "ambient space" $Y$.

*The trefoil knot, which may be parametrized by
$$
  G(\theta) = \frac{1}{1 + r_{2}\sin 3t}{(r_{1}\cos2\theta, r_{1}\sin 2\theta, r_{2}\cos 3\theta)},\ r_{1}^{2} + r_{2}^{2} = 1,
  $$
is the image of an embedding $g$ of the circle into $\Reals^{3}$. (The mapping $G$ is continuous in $\theta$, and injective as a mapping on the circle, since two values of $\theta$ map to the same location if and only if they differ by an integer multiple of $2\pi$. Since the circle is compact, the inverse mapping $g^{-1}$ is continuous.) By Point 2., any two embeddings of the circle into $\Reals^{3}$ are homotopic. Particularly, the mapping $f(x, y) = (x, y, 0)$ and mapping $g$ from the circle to the trefoil are homotopic. It turns out, however, that these mappings are not isotopic. To prove this, we would need an isotopy invariant of mappings that can distinguish the unknot from the trefoil knot.



The proper notion of topological equivalence depends on context, and quickly becomes subtle even in situations, such as a plane circle and a trefoil knot, that appear qualitative and simple.

*

*As topological spaces the circle and the trefoil are homeomorphic. In this sense, the circle and trefoil knot are "topologically equivalent".

*As mappings of the circle into $\Reals^{3}$, the circle and trefoil are homotopic, i.e., "deformable into each other through continuous mappings".

*As embeddings of the circle into $\Reals^{3}$, the circle and trefoil are not isotopic, i.e., "cannot be deformed into each other" through embeddings.

