Why does the compression criterion require homotopy relative to the boundary?

Question

The Compression Criterion states the following:

Let $(X,A,x)$ be a nested space triple and $$f: (D^n, \partial D^n, z) \longrightarrow (X,A,x)$$ a continuous map of space triples. Then the following are equivalent:

1. $f$ is triple homotopic to $const_x$.
2. $f$ is triple homotopic, relative $\partial D^n$, to a map with image in A.

The direction 2. $\Rightarrow$ 1. works by concatenating the given homotopy with one that contracts $D^n$ to $z$.

The direction 1. $\Rightarrow$ 2. is the interesting one: Here one can picture a cylinder standing upright with a horizontal slice corresponding to the homotopy at a point in time. The bottom face is $f$, the top face is $const_x$. The boundary maps to $A$. By blowing the bottom face into the cylinder as if it was made from rubber while fixing $\partial D^2 \times \{0\}$, and doing this all the way until the former bottom faces is contained in the cylinder's boundary again, we can find a homotopy from $f$ to a map with image in $A$, relative to the boundary.

A practically identical proof should show that:

3. $f$ is triple homotopic to a map with image in $A$.

implies 2. The picture is identical except that the top face maps into $A$, instead of being $const_x$. In particular a stronger version of the compression criterion would be that the following are equivalent:

1. $f$ is triple homotopic to $const_x$.
2. $f$ is triple homotopic, relative $\partial D^n$, to a map with image in A.
3. $f$ is triple homotopic to a map with image in $A$.

My question is the following: How come I have never seen this version of the compression criterion before, especially since for me, personally, this really helps with understanding relative homotopy? Or is this result not correct? In this case I must really be overlooking something.

Any help is greatly appreciated. Note that this question is a duplicate of Compression Criterion for $\pi_n(X,A,x_0)$. Why do we need homotopies $\text{rel} \ S^{n-1}$?, but since this is a 5 year old question with no answer and it seems like quite a relevant question, I am okay with possibly having this question deleted to get an answer to the original one.

Answer 1

The implications $2. \Rightarrow 3. \Rightarrow 1.$ are obvious. Let us prove $1. \Rightarrow 2.$

Let $H : (D^n,S^{n-1}) \times I \to (X,A)$ be a homotopy of triples from $f$ to a constant map $c : (D^n,S^{n-1})\to (X,A)$ with value $a_0 \in A$ (for our argument it is irrelevant wether or not $H$ is stationary on the basepoint $y$).

Define $M = \{(x,t) \in D^n \times I \mid \lVert x \rVert \ge \frac{2-t}{2} \}$ and $M' = \{(x,t) \in D^n \times I \mid \lVert x \rVert \le \frac{2-t}{2}\}$. These are closed subsets which cover $D^n \times I$. Define $$\bar H : D^n \times I \to X, \bar H(x,t) =\begin{cases} H(\frac{x}{\lVert x \rVert},2 - 2\lVert x \rVert) & (x,t) \in M \\ H(\frac{2x}{2-t},t) & (x,t) \in M' \end{cases}$$

1. $\bar H$ is well-defined and continuous since the definitions in line 1 and 2 agree on $M \cap M'$.

2. We have $S^{n-1} \times I \subset M$, thus for all $x \in S^{n-1}$ we get $\bar H(x,t) = H(x,0) = f(x)$.

3. For $x \in D^n$ we have $\bar H(x,1) \in H(S^{n-1} \times I) \subset A$ if $(x,1) \in M$ and $\bar H(x,1) \in H(D^n \times \{1\}) = \{a_0\} \subset S^{n-1}$ if $(x,1) \in M'$. Hence $\bar H(D^n \times \{1\}) \subset S^{n-1}$.

This shows that $f$ is homotopic relative $S^{n-1}$ to a map with image in $A$ (via the "compressing homotopy" $\bar H$). Clearly $\bar H$ is a homotopy of triples because it is stationary on $S^{n-1}$.

Let us finally come to the intuition to find $\bar H$. Look at the level $t$. The level set $D^n \times \{t\}$ can be decomposed in a closed ball $D^n_t$ with center $(0,t)$ and radius $\frac{2 -t}{2}$ and a closed spherical shell $A^n_t$ with center $(0,t)$ , inner radius $\frac{2 -t}{2}$ and outer radius $1$. Note that $D^n_0 = D^n \times \{0\}$ and $A^n_0 = S^{n-1} \times \{0\}$. We can identity $D^n_t$ with $D^n$ via "radial stretching", i.e. $(x,t) \mapsto \phi(x,t) = \frac{2x}{2-t}$. Define $\bar H(x,t) = H(\phi(x,t),t)$ on $D_t$. Similarly we can identify $A_t$ with $S^{n-1} \times [0,t]$ via $\psi(x) = (\frac{x}{\lVert x \rVert},2 - 2 \lVert x \rVert)$. Define $\bar H(x,t) = H(\psi(x,t))$ on $A_t$. Then on the common boundary sphere $C_t$ of $A_t$ and $D_t$ the map $\bar H$ is nothing else than $H_t \mid_{S^{n-1}}$ "compressed" to $C_t$.