Let $X$ and $Y$ be two unit circles in $\mathbb{R}^2$. $X$ is centered at $(1,0)$, while $Y$ is centered at $(-1,0)$.
Consider $A = X \cup Y$. Let $p \in Y$ be a point that is not $(0,0)$.
Why is $X$ a deformation retract of $A - p$?
Attempt: Intuitively, I can expand the hole in $Y$ and pushed both "arms" into the point $(0,0)$. But what would the retract $r: A - p \to X$ be exactly?
Also, the fact that $Y - p$ is homeomorphic to the real line via sterographic projection, and that the real line is contractible feels like I should be able to retract $Y-p$ to the point $(0,0)$.
Let $p = (-1 + \cos\psi, \sin\psi)$ for $\psi \in (-\pi, \pi)$. Consider $H: A\setminus\{p\} \times [0, 1] \to A\setminus\{p\}$ given by
$$H((-1+\cos\theta, \sin\theta), t) = \begin{cases} (-1 + \cos((1-t)\psi-t\pi), \sin((1-t)\psi-t\pi)) & \theta \in (-\pi, \psi)\\ (-1 + \cos((1-t)\psi+t\pi), \sin((1-t)\psi+t\pi)) & \theta \in (\psi, \pi) \end{cases}$$
and $H(x, t) = x$ otherwise. The first case retracts the bottom arc to $(0, 0)$ while the second case retracts the top arc to $(0, 0)$.