No retraction theorem in $\mathbb{R}^2$.
There is no retraction from the closed unit disc $\mathbb{D}=B(0,1)$ to its boundary $\mathbb{S}=\partial B(0,1)$.
In the proof , by contradiction, suppose $f$ is a retraction above, one choose two points $\alpha,\omega \in \mathbb{S} $, and let $$A=f^{-1}(\alpha),W= f^{-1}(\omega)$$ since they each intersects $\mathbb{S} $ at only one point, we can choose a subset $E \subset \mathbb{D} - A \cup W$ who is open and connected such that the closure of $E$ contains $\mathbb{S} $ ......
My query is why can we guarantee that $E$ is connected?
The proof contains a serious gap. Quote from the paper: "The two sets $A$ and $W$ may intersect themselves, so excising them may disconnect $\mathbb{D}$. However, because they each intersect $S^1$ at only one point, excising them leaves intact some subset of $\mathbb{D}$ whose closure includes the entirety of $S^1$. Let this set be denoted by $E$. Now E is connected."
It is not at all clear what $E$ should be. Is it simply $E = \mathbb{D} \backslash (A \cup W)$? This is not sure because "excising them (i.e. $A,W$) may disconnect $\mathbb{D}$." What else could be taken for $E$? Since it is claimed to be connected, it must be chosen as some component of $\mathbb{D} \backslash (A \cup W)$. But no indication is given which component.