How to prove the inclusion of an n-sphere in an (n+1)-disk is n-connected?

Question

Recall a continuous map $f:X\to Y$ is $n$-connected if for every $k\leq n$ and every commutative square $$\require{AMScd} \begin{CD} \mathbb S^{k-1} @>u>> X\\ @VVV @VVfV\\ D^k @>>v> Y \end{CD}$$ there exist:

1. a filler $\ell:D^k\to X$ such that $\ell|_{\mathbb S^{k-1}}=u$,
2. a homotopy $v\Rightarrow f\circ \ell \;(\text{rel }\mathbb S^{k-1})$.

Intuitively it seems to me the inclusion $\mathbb S^n\to D^{n+1}$ is $n$-connected. Indeed it "feels" possible to press the singular $k$-disk onto the boundary $n$-sphere of the $(n+1)$-disk and thus homotope it into the boundary.

$$\require{AMScd} \begin{CD} \mathbb S^{k-1} @>>> \mathbb S^n\\ @VVV @VVV\\ D^k @>>> D^{n+1} \end{CD}$$

Question. How to prove the inclusion $\mathbb S^n\to D^{n+1}$ is $n$-connected?

Answer 1

It is much easier to do this if the map from $S^{k-1}$ is constant. Then one has a pointed map $(S^k, *) \to (D^{n+1}, S^n)$ which you want to homotope into the boundary, relative to the point; because $D^{n+1}$ is contractible this is possible. The same homotopy (considered as a homotopy of $D^k$ rel the boundary) gives you the desired result.

To reduce to this, apply the homotopy extension lemma to $(D^k, S^{k-1} \times [0, \varepsilon])$ to a choice of homotopy rel $S^{k-1} \times \{0\}$ between the existing map $S^{k-1} \times [0, \varepsilon] \to D^{n+1}$ and a homotopy between the original map on the boundary and a constant map. The easiest way to do this is to start by homotoping the given map $S^{k-1} \times [0, \varepsilon] \to D^{n+1}$ to the map factoring as $S^{k-1} \times [0, \varepsilon] \to S^{k-1} \times \{0\} \to D^{n+1}$ ("undo" the first homotopy to make this constant), and then tracing out the given null-homotopy of $S^{k-1}$ inside the boundary. In the end, the map $S^{k-1} \times [0, \varepsilon] \to D^{n+1}$ is the same as the original map on $S^{k-1} \times \{0\}$, has image inside $S^n$, and restricts to a constant on $S^{k-1} \times \{\varepsilon\}$. Homotopy extension induces a homotopy of the entire map from $D^k$, relative to the boundary, of course.

Now using the fact that $D^k \setminus \left(S^{k-1} \times [0, \varepsilon)\right) \cong D^k$, you just need to run the first part of the argument to the disc of radius $1 - \varepsilon$.

Answer 2

The following seems relevant. It is from Topology and Groupoids, p. 304. The elegant formulation of the theorem and inductive proof is due to J.F. Adams in unpublished notes.

7.6.1 The following statements are equivalent for each $n \geqslant 1$:

$\alpha(n)$ Any map $S^r \to S^n$ with $r< n$ is inessential.

$\beta(n)$ Any map $S^r \to S^n$ extends over the disc $E^{r+1}$.

$\gamma(n)$ Let $B$ be pathconnected and let $Q$ be formed by attaching a finite number of $n$-cells to $B$. Then any map $$(E^r,S^{r-1}) \to (Q,B) $$ with $r < n$ is deformable into $B$.

The proof is by induction with $\gamma(1)$ being easy. It is then proved that $$\gamma(n) \Rightarrow \alpha(n) \Leftrightarrow \beta(n) \Rightarrow \gamma(n+1) $$ of which the last is the difficult step.