Is this animation a homotopy? I refer to the animation shown in this Wikipedia website of a homeomorphism. https://en.wikipedia.org/wiki/Homeomorphism
In fact, it is claimed that this animation depicts a continuous deformation (in fact it appears as though it illustrates a strong deformation retraction) between a torus and a mug. However I find this animation rather misleading in illustrating a homotopy. We shall consider the base of the mug to be a closed disk of radius 1 centred at the origin on the $xy$ plane, and assume that the standard topology be used at all times for purposes of this discussion. If one were to scrutinise the transition from a torus to a mug, which is to represent a homotopy $H:S^1\times S^1\times I\rightarrow M$ where $M = D^2\times 0\cup S^1\times I\cup K$ (where $K$ is the handle of the mug which is not of concern in this question), let $t^{'}\in I$ be the time $t^{'}$ where the torus is mapped to the solid defined by $D^2\times I$. Restricting $H$ to $S^1\times S^1\times [t^{'},1]$, which we denote by $H^{*}$, by assertion that $H$ is continuous then so is $H^{*}$. We transform $H^{*}$ linearly such that $H^{*}:D^2\times I\times I\rightarrow\mathbb{R}^3$ is a homotopy. Since $D^2\times I\cup K$and the torus is homeomorphic, with homeomorphism $p: D^2\times I\rightarrow S^1\times S^1$, the map $H^{*}\circ p\times Id_I: D^2\times I\rightarrow \mathbb{R}^3$ should also be continuous. With the standard topology endowed the standard tools of analysis are applicable. Consider any small neighbourhood centred at $(1,0,1,s)$ for $s\centernot=0,s\in I$. Wouldnt we be able to find a sufficiently small $\epsilon$ such that there exists such that no $\delta$ exists whereby whenever $||k-(1,0,1,s)||<\delta$, $||H(k)-H(1,0,1,s)||<\epsilon$?. Intuitively and loosely speaking, I cannot convince myself that the animation is continuous, as a chunk of solid is being vertically pushed down from $D^2\times I$.
Note that we modeled the mug to be of negliglble thickness. Since $D^2$ is connected we may change the argument to a mug with thickness without altering the purposes of this question. I visualise the deformation to motivare the construction of proposed homotopies and hence I feel this question is worth asking. Thanks in advance.
 A: You consider wrongly the mug as a surface, while it is clear from the figure that it is a solid body, and the displayed surface is the external surface of this body. So, if the handle of the mug would be removed, this surface could be defined as
$$\begin{align}
&\{(x,y,z)\mid x^2+y^2\le 10,\quad z=0\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2\le 9,\quad z=1\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2=10,\quad 0\le z\le 10\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2=9,\quad 1\le z\le 10\}\\
\cup\quad&\{(x,y,z)\mid 9\le x^2+y^2\le 10,\quad z=10\}
\end{align}$$ 
So, the first step of the animation is the homotopy for which the surface, at time $0\le t\le 1$ is:
$$\begin{align}
&\{(x,y,z)\mid x^2+y^2\le 10,\quad z=0\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2\le 9,\quad z=1+t(10-1)\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2=10,\quad 0\le z\le 10\}\\
\cup\quad&\{(x,y,z)\mid x^2+y^2=9,\quad 1+t(10-1)\le z\le 10\}\\
\cup\quad&\{(x,y,z)\mid 9\le x^2+y^2\le 10,\quad z=10\}
\end{align}$$ 
Can you be clearer about your concern with this homotopy?
