 Welcome to module 19. So now we will discuss a typical construction, a typical SDR, okay, strong deformation retract from which you can construct almost all well-known deformation retracts. As such there are not many ways of even constructing homotopes. So they all fall into a certain pattern, okay. Almost always things are happening in the domain itself, okay. So there are that kind of consideration is what I am going to do now, okay, to construct some of the typical strong deformation retract. So all these things we will carry out first of all in I cross I, okay. There slowly we will make it slightly little complicated but that is it. After that that can be used with slightly modifications and so on in many situations. So what I do in I cross I, I take the x-axis part and on the y-axis I take one cross I, okay. You put two the other two also. So instead of the entire boundary I am taking two of the sides, not opposite adjacent sides. So I cross 0 and one cross I. Then this subspace is strong deformation retract of I cross I. So like Sagnik says it will be intuitively clear but that is not where we live. We will write down a specific strong deformation retract. Even to do that there are many ways to do. So what I am going to tell you is a geometric way of doing this and this can be generalized to construct many things, okay. So here is it. What do I do? In I cross I, I have chosen this I cross 0 and one cross I. This is my subspace, okay. I want to get a strong deformation retract of the entire square onto the subspace A. So what I do? I take some point on the y-axis above this square somewhere. So here just for suitability I have chosen 0,2. That is not neat. I could have chosen this point or this point or far away 0, 200, no problem. From that point I am taking a view of the whole thing, okay. So first I see this one right from here and I can just keep, you know, if this is a transparent thing my eyes will see all the way to bottom here. So this line will tell me what I should do. So take a point here. I am going to push this point slowly to this point. This entire line 0, 0 to 0, 1 namely 0 cross I is going to push down to this point, okay. Then a point here, here we will go in here. This point we will go to this point, okay. So this is one part. You will have one single formula for this mark. But when you want further for a point here, we will be going to this point here, namely on the 1 cross I, the y-axis part, okay, this line. Then the formula will be different, okay. We have seen that, right. So I am going to push this one, this portion here. This point remains here, of course. There is nothing to push there. This way pushing means what? You do it at time t, at time 0 it is identity, nothing has happened. At time t, say t prime, this part has already pushed up to here. This point has come here, this point has come here, this point has come here, this point has come here and so on. Only this much is left out. To keep doing it at time equal to t equal to 1, only this segment and this segment will remain. So that is the homotopy. During this homotopy, points of A don't move. They are there, okay. Therefore, it is a strong definition, right. So this is geometry proof after this. Many books stop here. They don't write down. It is writing down is just for your sake. Namely, you get more confident. So I have done that. You can write down yourself. But you may not know how to write down such a thing. But these are just actually linear projections. So it is linear algebra, okay. So I have explained what it is and the p is 0, 2. You have to parameterize, take a point here, parameterize this portion from here to here, this portion, this line segment. Write down parameterization for this line. Use that parameterization as a homotopy, from this point to this point. At t equal to 0, it should be this point. t equal to 1, it should be this. You have to write a continuous way or say one formula up to here, from here to this point. Can you write down the coordinates of this point? What is this point? This point is the intersection of this horizontal line here and the line joining 0 to 1, 0. So you can write down. So this is just plain coordinate geometry, okay. So that is precisely what I have done here, okay. The map pushes the point arbitrary Z, whatever Z to RZ, okay. RZ is retract at the end result here, okay. Any Z here comes to RZ. The tip of the arrow will be the RZ. Point here will come here. Point here will also come here. So that is RZ, okay. The student of Topol is expected to appreciate this as a final proof and others are expected to swear at it in good faith. This is the way many books explain it, maximum explanation. They just go away out, okay. So if you want to feel that, if you want to do this kind of things on your own in a different situation, what is the way to learn? You have to, this is an easy method. Here you have to be very sure that you can write down the formula. So you have to write down, okay. So let me go through this for a while. I am doing it just for you. Let us prepare ourselves by working out a few such ideas into proofs so that we can appreciate such ideas as proofs later in many other situations. This is my, this is my contention. Of course, very first thing to verify is that the map result going to RZ is continuous. To see this, we divide the square i cross i into two parts by the line joining 1 comma 0 to 0 comma 2. That is what I told you, right. Let us say 1 comma 0 to this 0 comma 2. So this is one quadrilateral, that is a triangle. So it is divided into two parts like that. So one formula of R is this one. It pushes everything to x axis. The other formula pushes it into this y axis at the namely 1 cross i, okay. So here I have two different formulas. If Z equal to T, T prime is in the lower part, okay. Similarly T prime is less than equal to twice 1 minus 2, 1 minus T. Then RZ is given by this formula. On the other hand, if T prime is bigger than equal to 2 into 1 minus T, then RZ is this one. The first coordinate here is something. The second coordinate is 0. So it is in the x axis. Here the first coordinate is 1. Second coordinate is this one. Why this inequality, inequality, this inequality is give you the two divisions. This one will give you what I told you as the quadrilateral. This one gives you the triangle there. So on the line, on that line itself, both these formulas will agree. Namely the entire line goes to 1 comma 0. Therefore this is a continuous solution. After that homotopy between identity and R is just easy. Namely it is 1 minus S or T double prime Z plus T double prime RZ. In the case of, now you can appreciate why I did it for the case when the square minus center. I had to write down four different formulas. Once I have written down RZ, the homotopy is easy. That is only one way with strong homotopy, 1 minus T double prime Z. So T double prime because I have a T and T prime already used for M T double prime. So full formula for those T's in terms of RZ is written down. That is not necessary. That is a check here. Observe that the map S T, S little T pushes the square along the lines passing through 0 comma 2. So this is what all the time. The union of two line segments shown by the thick dotted broken lines in the image of segment T prime. All that I have told you. Okay. Once more we go back at T prime. This line, the top line becomes a broken line like this because this point is pushed here and this line segment is pushed to this slanted line here. At the final stage, this entire line is connected to this line, union this line segment. So this point goes here. Okay. Any questions here? Because now I am going to leave it here so that you verify all these formulas and then use this one to construct more complicated examples. So next time when you come before this one, you better check this one properly and understand this homotopy very carefully. Okay. Hopefully we will stop here.