I am watching these lectures on algebraic topology on youtube by NJ Wildberger. At around 9:00 they give an informal proof that there is no retraction $r:D\to S^1$ from the disk to its boundary.
I understand a retraction to be here a continuous mapping $r:D\to S^1$ such that $r(p)=p$ for all $p\in S^1$. The idea of the given proof, as far as a I understand it, is as follows:
- Suppose there is such a retraction $r:D\to S^1$.
- Consider continuous deformations of $r$ of the form $$p_t \equiv r\circ [\text{contraction by }(1-t)].$$ These maps will then be such that $p_0=r$, whereas $p_1:\{(0,0)\}\to r(0,0)\in S^1$ maps the center of the disk into some point on the circle.
- Consider a loop $\gamma$ on $S^1$. Consider the corresponding loops $p_t(\gamma):s\mapsto p_t(\gamma(s))$. Then $p_0(\gamma)=\gamma$ has winding number $1$, whereas $p_1(\gamma):s\mapsto r(0,0)$ is just a fixed point. Continuous deformations cannot change the winding number, thus there cannot be any such retraction.
I think I'm missing something with regards to this invariance of the winding number with respect to continuous deformations. I'm picturing the situation as considering how loops around $(0,0)$ with different radii get mapped by $r$.
Is there a more formal way to see why such deformations of $r$ cannot change the winding number? I could imagine a situation where $p_t(\gamma)$ is a loop on $S^1$, whereas $p_{t+\epsilon}$ is not a loop but rather a "slightly open" loop. I can imagine such opening to happen continuously. Why could such a behaviour not correspond to a continuous mapping $t\mapsto p_t(\gamma)$?