 So, let me begin with a result that we did in part 1 which is going to be useful again here. Take any subspace X, then the pair XA has homotopy extension property with respect to every space that is A to X is a co-fibration if and only if A is a closer subspace of X and the subspace Z which is A cross I union X cross 0 is a retract of X cross I. So, this is a more general statement in the previous lemma that we proved last time that this A was the boundary of the disk and X was the disk, so this is a more general statement. So, if A cross I union X cross 0 is retract of X cross I, then the pair X has a homotopy extension property and conversely, so this is the gist of the proposition here we are going to use it, there is no point in keeping recalling this and so on, you better read them on your own. So, this was the picture here, this is X and A is this triangle below, A cross I is this prism, this X cross I is the solid cylinder, the solid cylinder can be slowly deform leaving this portion here all the way the bottom unchanged, this entire thing collapse into this part. So, this is the meaning of that this X cross I can be strongly deformed to into X cross 0 union A cross I, this does not happen every time that is whenever the inclusion map is a co-fibration and conversely, ok. So, right now you can take this as a definition for co-fibration because they are equivalent. So, what we are going to do is that if X A is a relatively complex, then the inclusion map is a co-fibration. All that I am going to prove is that X cross 0 union A cross I is a strong deformation retract is just a retract of X cross I, that is enough sorry, it is not strong retract, this is just a retract of X cross I, ok. Actually in practice we have whenever we have constructed it is a strong deformation retract, this retract is enough, in the statement here you cannot push round even retract, retract is enough, ok. So, this is done very easily because now what we have is from previous lemma we have this thing the inductive step. If X is obtained from Y, by attaching K cells and X cross 0 union Y cross I is a strong deformation retract of X cross I, you do not need strong deformation retract is good enough, ok. So, you can put them together and get this wonderful result now, ok. It is enough to show that X cross 0 union X cross I is retract of X cross I, that is what we are going to do. Using the Able lemma successfully what you get the various skeletons there is a map R k from X k cross I to X k cross 0 union X k minus 1 cross I, right because X k is obtained from X k minus 1 by attaching K cells. So, now you define R this is true for every k define R from X cross I to X cross 0 union X cross I by the formula R whenever restricted to X k cross I namely when X belongs to X k comma t you define it as R naught composite R 1 composite R k. You take R k of that it will be in X k minus 1 cross I, then you take R k minus 1 and so on. Finally, you can take R naught to come back to the to the relative part A itself that is the meaning of this one, ok while while doing this X cross 0 is never disturbed it is identity here. So, A cross A cross I is also never disturbed for any of them so that is also identity. R is a automatically has this point set of a logical property why it is continuous because it is continuous restricted to each skeleton it is because it is just a when you restricted skeleton it is just a composite of finitely many continuous functions. So, essentially these infinite compositions are actually finite composition just like partitions of finitely infinite sums are actually finite sums there locally, ok. So, here is an exercise I will just I have used this one already so, but now I would like you to write down details that is why I have put this exercise here namely take any collection A of cells in X show that there is a unique sub complex k A of X such that A is contained inside k A the uniqueness comes unique smallest sub complex there is a sub complex A contained for k A, ok. So, k A is a sub complex A is just a collection of subsets if A is finite then k A is finite so, this is an important thing because after all see any collection of sets I have taken ok when you take the when you take the sub complex why it is finite there is some something you have to do here that X be countable write X as union of increasing sequence of subspaces X k this is not a k skeleton now by the way that the each SK is a finite sub complex. Countable sub complex you have to write it as union of finite sub complexes, ok. If you just take any collection of cells it will not be a sub complex so, that is why this first two parts are there. So, together the part 3 is the one which I have used actually and I have indicated that also but now we have to write a rigorous proof of that it is all, ok. So, that is the part for today I would not like to go ahead with another theme here so we will come to that one next time. Thank you.