Hi, thanks for the answer. Unfortunately, I still do not understand the way to get the homotopy. I have edited my post to include some more details. The picture that I added now is inspired by figure 51.5 in Munkres' topology book. But still, I don't understand how to get the homotopy. Please help. Thanks.

I won't be surprised if there is actually no way to construct the homotopy, i.e., the exercise in the OP wants the student to find the map which works as the desired homotopy by hit and trial. In which case, my question is pointless: as it is asking for something (the way to construct) that probably does not exist. I am not sure.

@Koro your comment doesn’t make sense. What do you mean by “get” the homotopy? I’ve discussed two different homotopies, which we have “gotten”

I don't know what's wrong with me. But I'm having too much difficulty in understanding this. So we want to define H on the square such that $H(s,,0)=\alpha(s), H(s,1)=\gamma\circ \beta\circ \delta^{-1}(s), H(0,t)=\alpha(0), H(1,t)=\alpha(1) \forall (s,t)\in I\times I.$ From the picture in your answer, I understand that we can define a map on $I\times I$ and needless to say that there are other ways one could cover up the square (like I tried to do in the image in my post by dividing the square by 3 vertical lines) but the problem is in ensuring that the homotopy properties are satisfied.

(Contd.) So suppose that we have a map defined, say using the image in your answer, let's call it G. How do I ensure that G satisfies properties satisfied with the desired H (i.e., behaviour of G at the end points etc.)?

@Koro The $G$ in my image is the same thing as the $H$ in your first question. This $H$ defines paths $H_t(x)$ that always stop and start at $F(0,0),F(1,0)$ which are the desired endpoints. In the image, you can see that by the way the lines begin and end where the alpha line begins and ends. The image is just for guidance, though, remember we are also applying $F$ to cast that square into $X$

Ahh I think now I am starting to understand it. Every path (marked by purple and red in the image) starts at $\alpha(0)$ and ends at $\alpha (1)$. But still, can you please elaborate a bit more on how to 'cast that square into X'? How to write H for a particular (s,t)?

@Koro I don’t understand your last question. You can already see how to write $H$ explicitly because you gave the formula in your post! By ‘cast the square into $X$’ that was just meaning, for any coordinate $(a,b)$ in the square we actually consider the point $F(a,b)$ in $X$

The problem is that I don't understand how 3st, s-st are coming in the definition of $H$ by Kevin. S answer. More importantly, how does the picture in your answer get me that? I'm sorry if my question sounds too trivial but it's not trivial to me.

@Koro my picture is not about Kevin. S’ answer, it’s about the first homotopy you referenced (“now for the first question”)

Ok. Then I have the same question: how is 4ts coming in the definition of H in the first homotopy by the use of picture? I have added an another example in my post, where I have the same problem-How to get H using the picture?

@Koro $4ts$ is just a choice that's convenient. It is a convenient interpolation from $0\to t$ as $x$ runs from $0\to1/4$. You are overthinking the details. All we need is something that works: there is no superior choice. The only thing you need to understand about a homotopy is (a) that it works and (b) how it actually transfers from one function to the other. As for the new example, you can blame your confusion on the fact that what is written is wrong! Notice that for $t=0$, $t\le 2s$ will always occur and the author claims that $H(s,0)=f(-s/2)$ makes sense, which it doesn't if $s>0$.

To demonstrate $f\simeq c_p\cdot f$ using that picture, we rather want: $$H(s,t)=\begin{cases}p&s\le\frac{1}{2}t\\f(s-(1-s)t)&s\ge\frac{1}{2}t\end{cases}$$

That was a typo in my post. I have edited (interchanged s and t in the numerator of argument of f) but still I have the same question :(.

Ohh. And how did you get that from the picture? Can you please explain? I understand that it works but the way you got it from the picture is what I don't understand. If I understand it, then I'm hopeful that I'll also understand the answer to my other question as well.

@Koro Sure. Imagine fixing $t$ (i.e. fixing a height on the picture). What does the path look like as $s$ varies? For some initial segment, we are stuck in the $p$-zone. This zone is $s\le\frac{1}{2}t$. Why specifically $\frac{1}{2}t$? Well, when $t=0$, I want the function $f$ - I do not want to be stuck in the $p$-zone. So at $t=0$ this zone should disappear. But at $t=1$ I want the path $c_p\cdot f$, which is stuck at $p$ for the first $[0,1/2]$, so $s\le\frac{1}{2}t$ suits these purposes. The line $s=\frac{1}{2}t$ is the leftmost line in the picture, delineating the triangle of this zone.

@Koro (Cont.) Ok, what do we do once we escape the $p$-zone? At $t=0$ we just want $f$, always. That is, we want $s$ to map to $f(s)$. But at $t=1$ we want $s$ to go to $f(2s-1)$. So, when we are outside this $p$-zone (the triangle) we need to interpolate between $s$ and $2s-1$. The picture suggests we do this linearly (those lines can be thought of as the images $(s,t)$ for fixed $s$ and varying $t$). The linear interpolation $s\to2s-1$ is exactly (algebra!) $(s,t)\mapsto s-(1-s)t$. Does it need to be linear? NO. But it's convenient.

(actually, please ignore my comment about "those lines can be thought of...", the lines are just guidance. They pretty much represent the paths from $f(s)\to(c_p\cdot f)(s)$ as $t$ varies)

I am afraid I still don't understand. "At t=1, we want s to go to f(2s-1)." I think that at t=1, we want f(s) from the picture. How do you get $\color{red}{2s-1}$ in f(2s-1)?

@Koro The composite path $c_p\cdot f$ takes values $f(2s-1)$ for $s\in[1/2,1]$ by very definition

Now I understand how they have defined $f(\frac{2s-t}{2-t})$ but not how you defined $f(s-(1-s)t)$ from the picture. $f( \frac{2s-t}{2-t})$ is there because at time t, by concatenation definition, we want $H_t$ to look like f(s), $0\le s\le 1$ so we must scale appropriately.