Here is the proposition and its proof (pg. 76 from the book named "Introduction to homotopy theory" by Martin Arkowitz):
And here is 1.4.2(3):
My questions are:
1- why the set $S$ took this form?
2- Why we needed 3 homotopies, $H^{''}, H^{'}, H$?
3- I looked at Appendix A (4 pages on point-set topology), but I could not understand specificlly why $H^{'}$ induces $H$?
Could anyone help me answer those questions, please?
I can't really tell if this is supposed to be in pointed spaces or not, since it doesn't specify. However, it uses $*$ which seems to be a base point in $Z$, so I suppose I'll assume they're pointed spaces, though it doesn't really matter.
Let's start with question 2. Why did we need 3 homotopies $H''$, $H'$, and $H$?
Well, first $H''$ isn't a homotopy, at least not in any strict sense. It's a map $A\times S\to Z$, and $S$ isn't the interval, so it's not really a homotopy.
The reason we need these maps however is that our goal is to construct a homotopy $H:CA\times I\to Z$, but we need to do this in stages, so we first construct $H'':A\times S\to Z$, and then $H':A\times I\times I \to Z$, and then $H:CA\times I\to Z$. Note that the domain is changing each time.
The answer to why we construct $H$ in this manner, requires an answer to questions 1 and 3.
Let's go backwards, since this is closer to how you might actually think of the proof. Let's start with question 3.
Question 3
We're given the data of two maps $h_0 : CA\to Z$ and $G:A\times I\to Z$ to start with, and we want to produce a map $H:CA\times I \to Z$ such that $H(i\times I) = G$, and $H_0=h_0$.
Well, we know that $CA$ is $(A\times I) / (A\times \{1\})$, and presumably the statement in appendix A should say something like the functor $-\times I$ preserves quotients (possibly explaining that $-\times I$ is a left adjoint, and therefore preserves colimits). I haven't checked Appendix A, but that's the point-set topology fact, being used in the proof.
Therefore $CA\times I := (A\times I/A\times 1\cup *\times I)\times I \cong A\times I \times I / (a,1,t)\sim (a',1,t)\sim (*,s,t)$, so by definition of the quotient space, a continuous map $H:CA\times I\to Z$ is equivalent to a continuous map $H':A\times I\times I \to Z$ such that $H'(a,1,t)=H(*,s,t)=*$ for any $a\in A$, $s,t\in I$.
Moreover, we also need $H'$ to satisfy that $H'(a,0,t) = G(a,t)$, and $H'(a,s,0)=h_0(\langle a,s\rangle)$, where $\langle a,s\rangle$ denotes the image of the pair $(a,s)$ under the quotient map $A\times I\to CA$.
Question 1
Now how can we take the data of $G$ and $H_0$ to define such an $H'$? Well, this is where $S$ comes in.
The point of $S$ (three sides of the square) is that we have a retraction $r:I\times I \to S$, so as long as we can define $H':A\times I\times I\to Z$ on the subset $A\times S$, we can extend it to all of $A\times I \times I$ by composing the part that we have defined $H'':A\times S\to Z$ with $A\times r : A\times I\times I\to A\times S$.
So now we're looking for a function $H''$ such that $H''(A\times r)(a,1,t) = H''(A\times r)(*,s,t)=*$, $H''(A\times r)(a,s,0)=h_0(\langle a,s\rangle)$, and $H''(A\times r)(a,0,t)= G(a,t)$.
However, $A\times S$ consists of points that are of the form $(a,s,0)$, $(a,0,t)$, or $(a,1,t)$ for some $a\in A$, $s,t\in I$. Thus we actually know what $A\times r$ does to most of these points, the only ones where we're not sure are the points $(*,s,t)$, but we know that $(A\times r)(*,s,t)$ still maps to something whose first coordinate is the base point.
Thus we want to find $H'':A\times S\to Z$ such that $H''(a,1,t)=*$, $H''(a,s,0)= h_0(\langle a,s\rangle)$, and $H''(a,0,t)=G(a,t)$. However this already defines $H''$ on every point of $S$, so this determines $H''$.
We just need to check that $H''(A\times r)(*,s,t)=*$ to make sure everything is valid.
Well, $(A\times r)(*,s,t)$ is either going to be of the form $(*,1,t)$, $(*,s,0)$, or $(*,0,t)$.
In the first case, everything maps to $*$, so we're ok. In the second case, $H''(*,s,0)=h_0(\langle *,s\rangle)=*$, since $h_0$ is a pointed map, and $\langle *,s\rangle$ is the base point of $CA$ for any $s$. Finally, in the third case $H''(*,0,t)=G(*,t)=*$, since $G$ is a pointed homotopy.
Thus $H''$ will induce the map $H'$ with the desired properties, which will in turn induce the desired homotopy $H : CA\times I\to Z$.
End note
For the unpointed case, the logic is identical, except that $CA=A\times I/A\times 1$ is the unreduced cone, and there's no worrying about the basepoints, we just define $H''(a,1,t) = H''(a,1,0)=h_0(\langle a,1\rangle)$ for all $t$ instead of defining it to be the basepoint of $Z$ for all $t$.