Let $M = \{(x, y, 0, 0) \in \mathbb{R}^4 ~|~ x^2 + y^2 = 1\}$ and $N = \{(0, 0, z, w) \in \mathbb{R}^4 ~|~ z^2 + w^2 = 1\}$ be subspaces of $S^3$. Construct a deformation retract of $S^3 \setminus M$ onto $N$, or show one does not exist.
Progress:
I was thinking something like the family of maps $t \in [0, 1]$ defined by $$ r_t(x, y, z, w) = \left (\sqrt{1 - t^2}x, \sqrt{1 - t^2}y, z, w \right ) $$ but the image is not on the sphere.
On
Hint. Every $p$ not on $M$ or $N$ is uniquely expressible as $p=\cos(\theta)m+\sin(\theta)n$ for points $m\in M$, $n\in N$ and convex angle $\theta\in[0,\frac{\pi}{2}]$. IOW all $p$ lie on an arc between points of $M$ and $N$.
BTW, fixing $\theta$ and allowing $m\in M,n\in N$ to vary yields a Clifford torus.
This hint illustrates the fact $S^3=S^1\ast S^1$ is a topological join of two Hopf circles. If we stereographically project $S^3$ using a pole through one of the circles, that circle is projected to an axis, and the other circle to a circle around that line. If we form a 2D cross-section through the axis, the other circle intersects it in two points, and if we draw all the arcs between the two points and the line, as well as draw all circles of constant $\theta$, we get bipolar coordinates.
Define $f_t : S^3 \setminus M \to S^3 \setminus M$ by $$f_t(x,y,z,w) = \frac{(tx,ty,z,w)}{\|(tx,ty,z,w)\|}.$$