 So, in the previous lecture, we talked about connected components and today we are going to introduce a notion of path connectedness which is like more intuitive notion of connectedness. So, let us just see what this is. So, definition. So, this path connected. So, let x be a logical space and we say x is path connected. If for any 2 points x and y in x, there is a continuous map gamma from 0 1 to x such that gamma of 0 is equal to x gamma of 1 is equal to y. So, if we take a picture. So, auto logical space x might be like this x and this y and then there is some gamma which could be very complicated which connects x to y. So, the first thing we want to prove is that first result we want to prove is a path connected space, path connected space is connected. So, being path connected is a stronger notion than being connected. So, proof. So, let us assume that x is path connected, but not connected. Then since it is not connected, we can write x as a disjoint union u disjoint union v where u and v are non-empty open subsets. So, since they are non-empty, let x be a point in u and y be a point in v. So, then as x is path connected, there is a continuous map gamma from 0 1 to x such that gamma of 0 is equal to x and gamma of 1 is equal to y. So, now we simply take. So, the inverse image. So, then 0 1 we can write it as gamma inverse u disjoint union gamma inverse v. Since u and v are disjoint, the inverse images are also going to be disjoint. And 0 is in gamma inverse u because gamma of 0 is equal to x which is u and 1 belongs to gamma inverse of v. And both these are open. So, thus we have written 0 1 as a disjoint union of non-empty open subsets. So, they are open because gamma is continuous and therefore the inverse image of an open subset is open. So, this contradicts the connectedness of 0 1. So, a path connected space is connected. So, this completes the proof. So, let us see some examples. So, 0 1 all these nice examples that we know are all path connected. So, I mean, if we take the interval 0 1. So, then any two points a and b, we can join it by a straight line. So, we can define gamma of t is equal to 1 minus t times a plus t times b. So, this is then gamma is a continuous map from 0 1 to this interval 0 1 such that gamma of 0 is equal to a and gamma of 1 is equal to b. So, any two points can be connected by this path. So, the same works for the same gamma works for shows that r and r n are also connected or path connected. So, for r n we can just take two vectors and we can join them by a straight line once again. So, gamma is 0 1 to r n and this is gamma of t is equal to let us say we have two vectors a plus t times b. So, let us look at s 1. So, s 1 is this circle over here. So, once again given any two points let us say this is at an angle theta 1 and let us take another point here which is at an angle theta 2. So, this a and this b. So, then we can connect. So, the point is we can connect using a path which starts at a and goes all the way here right. So, but if you were to write it explicitly so, then we can define gamma of t is equal to e to the power 1 minus t times theta 1 plus t times theta 2. So, we have defined this map gamma t and we need to show that this is continuous. So, we will write it as a composite of two continuous maps right. So, we have this map from 0 1 to r which is t goes to 1 minus t times theta 1 plus t times theta 2 and then we will take the map from r to c which is x goes to e to the power x. So, this map is continuous and therefore, this composite is continuous. So, let us try to see that the spheres ok. So, this is s 1, but how about the spheres right. So, let us just do the example for s 2. So, for s 2 let us say this is the equator this is the north pole. So, let us call it n and this is the south pole this is s. So, if p is any point which is not equal to the south pole then we can what we can do is we can connect we can take this point p we can join the by the straight line right. So, first we consider this path gamma 1 t that is equal to t times n or rather 1 minus t times n plus t times p right, but now the straight line is not going to lie on the sphere. So, p is not equal to s that is important. So, we have to make this lie on the sphere. So, for that note that 0 does not lie on the straight line right the origin. So, maybe I should write 0. So, then this implies that the norm of gamma 1 t is a continuous map from 0 1 to r right and therefore, we can just define this map gamma 1 t by this from 0 1 and the image now will land inside the sphere s 2. So, what is this map? This map is t goes to gamma 1 t divided by norm of gamma 1 t ok. This is continuous. So, this is actually this goes to positive. This is continuous because we are taking a continuous map and dividing it by another continuous map right. So, each of the coordinates of gamma 1. So, if I were to write this as gamma 1 1 t is divided by norm right. Each of these coordinates is a because the denominator is never 0 ok. So, this shows that. So, this shows that n can be connected p using a path. Now, but we want to connect n to s the south pole also. So, for that I mean we can just take two paths this join them. So, we can take this point. So, first we can connect n to this point and then in the same way that we use constructed this path from n to a point p we can construct a path from the south pole to the point p. So, this gives a combined path from the north pole to south right. So, this shows that. So, this shows that s 2 is. So, how to combine two paths is something which we will see precisely in a minute. So, and same proof can be slightly modified. So, a slight modification of this proof also shows that s n right a slight modification shows that s n is path connected for all n gradient. In fact, this proof is better because then we do not have to worry about x goes to e to the power x is continuous right. We can just do away with that ok. So, we have seen some examples of path connected spaces some nice examples where we actually make some pictures and intuitively we imagine ok. So, but now let us continue with our discussion. So, as before ok. So, before we continue like we can also ask ourselves. So, we have seen many spaces so far. So, what about the path connectedness of those. So, let us what the other examples we have seen we have seen g l n r we have seen m n r we have seen the orthogonal groups special orthogonal groups we have seen the unitary matrices s u n right. What about connectedness of these right. So, g l n r let us is not even connected because we have this determinant map from g l n r minus 0 right which is surjective because we can just take minus 1 determinant of this is minus 1 and of course, determinant of the identities plus 1 right. So, therefore, r minus 0 which is disconnected we can write it as u disjoint here in v where u is this half open interval and v is this half open interval right and this is the point 0 which you are missing out. So, when we take. So, the determinant is a surjective map from g l n r to in fact we can just put any lambda over here lambda on 0. So, we get this surjective map. So, this implies that determinant inverse of u disjoint unit determinant inverse of v both are non empty gives. So, we can write g l n r as the disjoint unit of 2 non empty open subsets. So, therefore, g l n r is not connected right. So, but on the other hand we can ask what about g l n r plus this equal to those matrices A in or this is just equal to determinant inverse of matrices with positive determinant. So, we will show that later. So, m n r is just r n square and this is a vector space. So, since r n is path connected that same proof gives that m n r is path connected. Once again the orthogonal groups are not path connected for the same reason because when we take determinant the determinant it lands inside this set plus minus 1 right. And this plus minus 1 is just one disjoint unit minus 1 and once again the determinant is surjective. So, when we take the inverse image of both these open subsets. So, we get that shows that o n is not connected. So, it is important to say that the determinant is surjective because if you look at I mean if you look at g l n r plus. So, let me just write g l n r right. So, when we look at the determinant map at the level of g l n r plus the image lands inside minus infinity comma 0 disjoint unit 0 comma infinity right. But when we take the inverse image of the determinant, determinant inverse of minus infinity comma 0 is empty over here right. So, therefore this does not show that g l n r plus is disconnected. So, what I am trying to emphasize is it is important to say that this map is surjective or at least it meets the components of more than one component of the target space more than one connected component of the target space ok. So, but S o n. So, o n is not path connected is not connected yeah, but S o n is path connected right similarly u n and S u n. So, these are these three examples are very interesting, but they are also difficult to prove and to prove these we need the notion of quotient apology which we will introduce later on in this course. So, we will towards the end of this course we will prove that these three spaces are path connected. And I should say that one should emphasize this result because it is very interesting because S o n I mean is defined as it has a very complicated definition a transpose is equal to identity and determinant of a is equal to 1 right. So, with this complicated definition a priori if one just looks at the definition it is not all clear if it is path connected, but yes we can join any point any two points in S o n with a path which is completely inside S o n ok. So, now let us continue as before an equivalence relation ok. So, this time. So, we define the relation as follows. So, we say that. So, let. So, we define x is equal to y if there is a path from continuous map from 0 1 to x such that gamma of 0 is equal to x and gamma of 1 is equal to y right. So, this is new equivalence relation and let us check that. So, let us check this defines an equivalence relation right. So, we need to check three things. So, first we need to check that x is equal to x right. So, this is easy. So, we just take take the constant path right. So, in other words gamma from 0 1 to x and gamma of t is equal to x for all t right and the constant path is continuous. So, the second is if x is equal to y then y is equal to x right. So, suppose since x is equal to y there is a path gamma such that gamma of 0 is equal to x and gamma of y gamma of 1 is equal to y right. So, then take the take gamma 1 define gamma 1 of t is equal to gamma of 1 minus t right. So, gamma 1 is the composite of 0 1 to 0 1 this is t goes to 1 minus t which is obviously continuous and then here we have gamma right. So, gamma 1 is a path from. So, gamma 1 is a path ok. So, basically what if this is gamma this x this is y. So, gamma is going like this then gamma 1 traces it in the opposite the same path in the opposite direction. So, in particular gamma and gamma 1 have the same image inside x they are this they are different maps from 0 1 to x, but they have the same image ok. And the third thing we need to check is if x is equal to y and y is equal to x then x is equal to z right. So, here we will use a theorem that we learnt sometime back it is about how to check continuity of a map by restricting it to 2 close subsets right. So, let since x is equal to y then x is we have 2 we have path gamma 1 from 0 1 to x a continuous map such that gamma 1 of 0 is equal to x and gamma 1 of 1 is equal to y. And since y is equal to z we have this path gamma 1 sorry this is gamma 2 gamma 2 of 0 is equal to y and gamma 2 of 1 is equal to z right. So, basically this x this y and this z right. So, here we have gamma 1 and here we have gamma 2. So, now we define this map from h ok first we define a map from h 1 0 half x. So, h 1 of t is equal to gamma 1 of 2 t and then we define a map h 2 from half gamma 1 to x by h 2 of t is equal to gamma 1 of 2 t minus gamma 2 sorry of 2 t minus 1 ok. So, let us check that h 1 of 2 t is equal to 1 of half is equal to h 2 of half right. So, this is this is equal. So, h 1 of half is gamma 1 of 1 which is equal to y and h 2 of half is gamma 2 of 0 which is also equal to y right therefore, they are equal right. So, now this what does this mean this means that on the interval 0 1 this half right let us call this set A and let us call this set B. A is the closed interval 0 half and B is the closed interval half 1 right. So, this to x we have defined this map. So, this map is h 1 on this 0 half and it is h 2 on half gamma 1 on B right. So, this defines a map h and this map is I mean this actually defines a map because both h the only point of intersection. So, h 1 is a map on A and h 2 is a map on B and both h 1 and h 2 agree on the intersection in this case the intersection is just the point half right. So, therefore, it defines a map we get a map of sets right now h from 0 1 to x. So, the question is is h continues. So, now recall a theorem right. So, h is continuous. So, since right a comma B contained in 0 1 are closed subsets closed subsets right. So, h from 0 1 to x is continuous if and only if h restricted to a and h is restricted to B are continuous. So, obviously, A and B have the subspace topology right, but h restricted to A is exactly h 1 right and what is h 1 h 1 is h 1 is a composite. So, first we go from 0 half to 0 1 t goes to 2 t right and then we apply gamma 1 to x and gamma 1 is continuous and this map is continuous. So, therefore, their composite is continuous and similarly h restricted to B is equal to h 2. So, what is h 2? First we go from half comma 1 to 0 1 by t goes to 2 t minus 1 and then we compose with gamma 2. So, gamma 2 is continuous this map is continuous the first this t goes to 2 t minus 1 right because we have seen that all polynomials in the coordinates are continuous. So, therefore, this composite is continuous right. So, this implies that h 1 restricted to A and h restricted to A and h restricted to B are continuous this implies that h is continuous right. So, more over h of 0 is equal to. So, h is defined to be h 1 on the interval 0 half right. So, h of 0 is equal to h 1 of 0 which is equal to gamma 1 of 0 which is equal to 0 right and h of 1 is equal to. So, h is defined as h 2 on the interval B. So, it is h 2 of 1 which is equal to gamma 2 of 1 which is equal to z. So, thus h is path from x to z. So, this shows that x is equal to z ok. So, and this also shows if I have topological space x and if I can join 2 points x and y with the path and y and z with the path then I can join x and z with the path which is precisely what we used over here in the example of the sphere. So, we can join the north pole to any point P and we can join the north pole to this point P right and then P we can join to this point S right. So, therefore, we can join the north pole to S. So, this defines an equivalence relation. So, the equivalence classes the path components and similar to the proposition about connected components that we proved last time we have this proposition about path components every path connected subset. So, ok. So, we can write x as a disjoint union of x i's. Now x i's are path components every path connected subspace of x is contained some x i. In fact, it is contained in a unique x i right and 2 every each x i is path connected. So, both these will prove that x i maximal subspaces. So, notice that earlier we had a third point which said that when we looked at the connected components we had that the connected components are also closed. But this is not true for path components ok. So, we will see a counter example in the next lecture. So, remark unlike. So, maybe I should make this in red. So, unlike the connected components which were closed in x the path components need not be closed. So, we will end this lecture here.