If $f$ restricts to the identity map on the boundary $S^1,$ then $f$ must be onto (revisted).

Question

Here is the question I am trying to understand its solution:

Show that if $f: D^2 \to D^2$ restricts to the identity map on the boundary $S^1,$ then $f$ must be onto.

Here is the solution given here Continuous function from the closed unit disk to itself being identity on the boundary must be surjective? but I do not understand: 1- What exactly the author is doing in the following part of the solution:

"Now compose with the map $D^2 \to S^1, x \mapsto x/\|x\|$. This defines a retraction $D^2 \to S^1$. But that's silly, as if there were such a retraction, the map $\pi_1(S^1) \to \pi_1(D^2)$ induced by the inclusion would be an injection, and it's not."

2-And what is the importance of the Alexander trick mentioned in the comments for the solution and what is the importance of the automorphisms of the unit disc in the solution.

3- where is the retraction that we will get?

3- why the map $\pi_1(S^1) \to \pi_1(D^2)$ induced by the inclusion would be an injection, and why it's not? is this because of the following post There is no retraction map from unit disk to its boundary.(2) ?

Could someone help me in understanding this please?

Answer 1

1- In that statement, the author is constructing a map that projects each point in $D^2-\{0\}$ to its nearest point on the boundary ($x/||x||$ has magnitude 1, and so the map just increases the magnitude of each point in $D^2$ to 1). They then make the point that such a retraction, were it to exist, would imply that the induced map $\pi_1(D^2)\to\pi_1(S^1)$ between the fundamental groups would be injective (this is an early result in algebraic topology). However, $\pi_1(S^1)\cong\mathbb{Z}$ and $\pi_1(D^2)$ is the trivial group 1, and we cannot possibly have an injective map $\mathbb{Z}\to1$.

2- They discuss the Alexander trick as a part of a follow up question about the Brouwer fixed point theorem, so not sure that matters to the part you're interested in.

3- The retraction cannot exist; this is the point of the proof, and demonstrates that the function in question must be surjective.

Answer 2

Now compose with the map $D^2 \to S^1, x \mapsto x/\|x\|$.

This does not work, the map is not defined for $x = 0$. We have to consider the map $r : D^2 \setminus \{0\} \to S^1, r(x)) = x/\|x\|$. How can we use it?

If $f$ is not onto, there exists a point $p \in D^2$ such that $p \notin f(D^2)$. Since $f$ is the identity on $S^1$, we must have $p \in \mathring D^2 = \{ x \in \mathbb R^2 \mid \lVert x \rVert <1 \}$. There exists a homeomorphism $g : D^2 \to D^2$ resricting to the identity on the $S^1$ such that $g(p) = 0$.

Then consider the map $$F : D^2 \stackrel{gf}{\to} D^2 \setminus \{0\} \stackrel{r}{\to} S^1 .$$ This map is a retraction. Denoting by $i : S^1 \to D^2$ the inclusion map, we get $r \circ i = id$ and thus $r_* \circ i_* = id$ for the induced maps on fundamental groups. This means that $i_* : \pi_1(S^1) = \mathbb Z \to \pi_1(D^2) = 0$ must be injective, which is nonsense.

The existence of $g$ is not proved, but we can use the the links in the comments to find such a map. I think we do not need the Alexander trick to do it, we can use other methods. But that is another question.