In Hatcher Chapter 4, he defines the relative homotopy group $\pi_n(X,A, x_0)$ as maps $(I^n, \partial I^n, J^{n-1})\to (X, A, x_0)$ up to homotopy through such maps.
$J^{n-1}$ is defined as the closure of $\partial I^n\setminus I^{n-1}$, where $I^{n-1}$ is the facet of $I^n$ with $x_n=0$, namely $I^{n-1}\times\{0\}\subset I^n$.
Then he says that collapsing $J^{n-1}$ converts $(I^n, \partial I^n, J^{n-1})$ to $(D^n, S^{n-1}, s_0)$, so the group is the same as maps $(D^n, S^{n-1}, s_0)\to (X, A, x_0)$ up to homotopy of such maps.
The pictures of how $(D^n, S^{n-1}, s_0)$ is a quotient space of $(I^n, \partial I^n, J^{n-1})$ are clear to me. I don't think I need more pictures in this problem: I want to find the explicit quotient map.
Would someone help me find the explicit quotient map?
Background:
I don't know if this is shared by many people but I don't feel comfortable at this stage of my learning to be satisfied with just pictures. I don't think I have that right in some sense, because I don't think I would be able to give the explicit map if needed be, like in this case.
For context, I found (with a great help from this forum) the equivalent solution for the absolute homotopy groups: maps $(I^n, \partial I^n)\to (X,x_0)$ can be seen as maps $(S^n, s_0)\to (X, x_0)$ and vice-versa. All we need to show is an explicit quotient map $(I^n, \partial I^n)\to (S^n, s_0)$, as follows:
First we map $(I^n, \partial I^n)\to (D^n, S^{n-1})$ by first considering a change of coordinates on $I^n$, so that $I^n = [-1,1]^n$.
We take $x\mapsto \frac{x}{||x||}\cdot\displaystyle\max_{1\leq j\leq n}|x_j|$, and $0\mapsto 0$. It is a straighforward exercise to show this is a homeomorphism in fact (I first showed continuity, then bijectivity, then being a map from Compact to Hausdorff gives it to be closed and thus a homeo).
We proceed by finding a quotient map $f:(D^n, S^{n-1})\to (S^n, s_0)$ where it is injective everywhere but at the boundary $S^{n-1}$, which gets all mapped to $s_0$. This provides the homeomorphism between $(D^n/S^{n-1}, s_0)$ and $(S^n, s_0)$.
The map is, considering $D^n$ in $\mathbb{R}^n$ and $S^n$ in $\mathbb{R}^{n+1}$:
$d:\mathbf{x}\mapsto \left(2\sqrt{1-||\mathbf{x}||^2}\mathbf{x}, 1-2||\mathbf{x}||^2\right)$
(It is again a straightforward exercise to show this is the map we claim it to be. We first show continuity, then surjectivity, then since it is a map from Compact to Hausdorff we get it is closed and thus a quotient map. The final step is showing it is injective everywhere except at the boundary of the disk, all mapping to $s_0$).
By the universal property of quotient spaces, the above quotient map and the homeomorphism from n-cube to n-ball, we conclude there is a natural bijection between maps $(S^n, s_0)\to (X,x_0)$ and maps $(I^n, \partial I^n)\to (X, x_0)$. We may also show homotopic maps goes to homotopic maps, so we naturally identify $[I^n, \partial I^n; X, x_0]$ and $[S^n, s_0; X, x_0]$.
The analogous result for $(I^n, \partial I^n, J^{n-1})$ and $(D^n, S^{n-1}, s_0)$ is what I am looking for.
As seen from the above, I don't mind that the quotient space is built in multiple steps. I just want something that is explicit at every step, and not just pictures.
What I did so far:
Following the map described in the background section above $\phi:I^{n-1}\to D^{n-1}$, we consider a point in $I^n = [-1,1]^n$ as $(\mathbf{x}_L, x_n)$, with $\mathbf{x}_L=(x_1, \dots, x_{n-1})$, and build the map
$\Phi: (\mathbf{x}_L, x_n)\mapsto \left(\phi(\mathbf{x}_L), \frac{x_n}{2}\right) + (\frac{1}{2}\mathbf{1}, 0)$
which is a homeomorphism of $I^n$ to $D^{n-1}\times [0,1]$, the cylinder with height equal to the radius on the base. This cylinder contains a semi-ball with same base $D^{n-1}\times\{0\}$.
We can further apply the map defined on such cylinder
$(\mathbf{x}_L, x_n)\mapsto \min(||\mathbf{x}_L||, x_n)\cdot\dfrac{\mathbf{x}}{||\mathbf{x}||}$, if $||\mathbf{x}_L||\cdot x_n\neq 0$
Else $(\mathbf{x}_L, x_n)\mapsto (\mathbf{x}_L, x_n)$.
This maps the cylinder homeomorphically to the semi-ball $\{(\mathbf{x}_L, x_n): ||\mathbf{x}||\leq 1, x_n\geq 0\}$.
So now this could theoretically make the problem simpler. In these maps it's easy to see $J^{n-1}$ gets mapped to the half $n-1$-sphere, and the boundary gets mapped to boundary, so it would be sufficient to show a quotient map of the half-disk to a full disk which is injective everywhere but at the half $n-1$-sphere, which gets all mapped, say, to the south pole.
Another Partial Result:
We can build a quotient map on the referred semi-ball inspired by the quotient map $d:D^n\to S^n$ above. It will be a map $\tilde{d}$ which assign the disk at height $x_n$ of the semi-ball to the sphere of radius $1-x_n$ of $D^n$:
$\tilde{d}:(\mathbf{x}_L, x_n)\mapsto (1-x_n)\cdot d\left(\dfrac{\mathbf{x}_L}{\sqrt{1-x_n^2}}\right)$, where the argument of $d$ goes over an $n-1$-disk of norm $1$ in $\mathbb{R}^{n-1}$.
This map is a quotient map, identifying points in the same copy of $S^{n-1}$ at height $x_n$, for $x_n\geq 0$ (we first show continuity, then surjectivity, then being a map from Compact to Hausdorff shows it is a quotient map). These identifications are deduced from the identifications of $d$.
This is now an intermediary quotient space which is homeomorphic to the n-disk. The face $I^{n-1}$ of the cube went to the boundary of this n-disk, where we want it to go, but $J^{n-1}$ went to the line segment given by $(\mathbf{0}, -t)$, $0\leq t\leq 1$.
Final map:
We consider the following map on this disk $D^n$: we map the radial segments of boundary points with $x_n\geq 0$ to the segment between such boundary points and $(\mathbf{0}, -1)$ linearly.
For points with $x_n\leq 0$, we look at the horizontal segments between the boundary points with $x_n\leq 0$ and their projection onto the $x_n$-axis. The map will take such segments to the segments between such boundary points and $(\mathbf{0}, -1)$ linearly as well.
It is an immediate characteristic of this map that it fixes the boundary points. A formula for it is:
$(\mathbf{x}_L, x_n)\mapsto \left(\mathbf{x}_L, (||x||)\cdot\left( \dfrac{x_n}{||x||}\right)+(1-||x||)\cdot(-1)\right)$ if $0\leq x_n\leq 1$,
$(\mathbf{x}, x_n)\mapsto \left(\mathbf{x}_L, \left(\dfrac{||\mathbf{x}_L||}{\sqrt{1-x_n^2}}\right)\cdot(x_n) +\left(1-\dfrac{||\mathbf{x}_L||}{\sqrt{1-x_n^2}}\right)\cdot(-1) \right)$ if $-1 < x_n \leq 0$,
$(\mathbf{0},-1)\mapsto (\mathbf{0}, -1)$.
This is a map that is continuous because it is continuous on the two closed sets $\{x_n\geq 0\}$ and $\{x_n\leq 0\}$, and agree on their intersection. The last piece is to establish continuity on $(\mathbf{0},-1)$, a removable singularity.
It is also surjective by construction. As a map from Compact to Hausdorff, it is a quotient map. It is easy to show that it is injective everywhere except at the segment $(\mathbf{0}, -t)$, $0\leq t\leq 1$, which gets all mapped to $(\mathbf{0},-1)$.
This final map is therefore a quotient map from the n-ball to itself, and thus makes a homeomorphism between $D^n/\{(\mathbf{0}, -t),\; t\in I\}$ and $D^n$.
Finally, the composition of all these maps described above establishes a quotient map $(I^n, \partial I^n, J^{n-1})\to (D^n, S^{n-1}, s_0)$. This shows the two formulations of $\pi_n(X,A, x_0)$ are equivalent.