Let the Möbius strip be $\mathcal M=[0,1]\times [0,1]/\sim$ where $(0,t)\sim (1-t,1)$. Let $$A=\{(x,y)|(x-1/2)^2+(y-1/2)^2=1/9\}\cup (\{1/2\}\times [0,1/6])\cup (\{1/2\}\times [5/6,1])$$ Show that $A/ \sim$ is homeomorphic to $S^1\cup ([-1,1]\times \{0\})$. Also, prove that $\mathcal M\setminus \{(1/2,1/2)\}$ deformation retracts to the subspace $A/ \sim$.
Although I have an intuitive idea of the homeomorphism, I can't see any way to write an explicit map. And, I have no clue about how $\mathcal M\setminus \{(1/2,1/2)\}$ can deformation retract to $A/ \sim$.
Please note that clues and intuitive explanations are welcome, but it would be very helpful if they are accompanied with explicit maps.
I believe $\sim$ should be defined by $(t,0)\sim(1-t,1)$ in order for the rest of your question to make sense.
For the first part of the question, we will show that $A/{\sim}$ is homeomorphic to $Y=S^1\cup(\{0\}\times[-1,1])$ because the desired result would then follow from applying a rotation by $\pi/2$ radians.
The equivalence relation $\sim$ identifies the point $(1/2,0)$ with $(1/2,1)$. We define a continuous map $f:A\to Y$ explicitly as follows. $$ f(x,y)=\begin{cases}3\left(x-\frac12,y-\frac12\right) &\text{ if }\left(x-\frac12\right)^2+\left(y-\frac12\right)^2=\frac19\\ \left(x-\frac12,6-6y\right) & \text{ if }(x,y)\in\{1/2\}\times[5/6,1]\\ \left(x-\frac12,-6y\right) & \text{ if }(x,y)\in\{1/2\}\times[0,1/6] \end{cases} $$ It is easy to check that $f$ is continuous and satisfies: \begin{align}\forall (x,y),(x',y')\in A,\ (x,y)\sim(x',y') \Leftrightarrow f(x,y)=f(x',y')\tag{1}\end{align} Thus, by the universal property, there exists a continuous map $\tilde{f}:A/{\sim}\to Y$ such that $f=\tilde{f}\circ p$, where $p:A\to A/{\sim}$ is the quotient map. You can check that $\tilde{f}$ is bijective. Since $A/{\sim}=p(A)$, we can conclude that $A/{\sim}$ is compact, so that $\tilde{f}$ is a continuous bijection from a compact space to a Hausdorff space and hence a homeomorphism.
The red part of the following picture represents $A/{\sim}\subset \mathcal M\setminus\{(1/2,1/2)\}$
The yellow arrows indicate how the deformation retraction works. To simplify the expression, we will first define a map $G((x,y),t):(\mathcal M\setminus B_{1/3})\times I\to \mathcal M\setminus B_{1/3}$ that shrinks the two sides to $A/{\sim}$. Note that $B_{1/3}=\{(x,y)\mid (x-1/2)^2+(y-1/2)^2<1/9\}$.
$$G((x,y),t)=\begin{cases} (1-t)(x,y)+t(\frac12,y) & \text{ if }y\in[0,1/6]\cup[5/6,1]\\ (1-t)(x,y)+t\left(\sqrt{\frac19-(y-\frac12)^2}+\frac12,y\right) & \text{ if }y\in[1/6,5/6],\ x\ge\frac12\\ (1-t)(x,y)+t\left(-\sqrt{\frac19-(y-\frac12)^2}+\frac12,y\right) & \text{ if }y\in[1/6,5/6],\ x\le\frac12\\ \end{cases}$$ Clearly, $G$ is well-defined and continuous. Now, we define $H:(\mathcal M\setminus \{(1/2,1/2)\})\times I\to\mathcal M\setminus \{(1/2,1/2)\}$ as follows. $$H((x,y),t)=\begin{cases} (1-t)(x,y)+t\frac{(x-\frac12,y-\frac12)}{3\Vert(x-\frac12,y-\frac12)\Vert}+t(\frac12,\frac12) & \text{ if }(x-1/2)^2+(y-1/2)^2\le 1/9\\ G((x,y),t) & \text{ otherwise} \end{cases}$$