proving $(\mathbb{B}^2 \times \{0 \}) \bigcup (\mathbb{S}^1 \times [0, \infty))$ is a retract of $\mathbb{R}^3$

Question

I don't know how to solve this. I tried constructing a retraction but nothing comes to mind. Can someone guide me through this and if possible explain the intuition behind solving this kind of problem? Thank you.

Answer 1

My intuition here would be that $(\mathbb B^2 \times \{0\}) \cup (\mathbb S^1 \times [0,\infty))$ is homeomorphic to $\mathbb R^2 \times \{0\}$ via a homeomorphism $f: (\mathbb B^2 \times \{0\}) \cup (\mathbb S^1 \times [0,\infty)) \to \mathbb R^2 \times \{0\}$ fixing $\mathbb B^2 \times \{0\}$. If we can extend $f$ to a homeomorphism $f: \mathbb R^3 \to \mathbb R^3$, then the desired retraction can be achieved as $f^{-1} \circ p \circ f$, where $p: \mathbb R^3 \to \mathbb R^2 \times \{0\}$ is the projection retraction. It may be somewhat tedious to construct this extension of $f$ explicitly, but it seems intuitive that it can be done.