 So, welcome to this lecture. In the previous lecture we saw that G L and R plus is path connected. So, let us begin this lecture by showing that proposition G L and C is path connected. So, to show that G L and C is path connected. So, given a matrix A in G L and C we will find joining A to identity. So, we have to find path in G L and C. So, let us just look at this straight line path. So, let consider this gamma t this is t times identity plus 1 minus t times A ok. So, now note that if t is a complex number then gamma t is actually a matrix in. So, this i n is n times n identity matrix then gamma t is a matrix in M and C. It is an n cross n matrix with complex entries right. So, what we need? So, we need to find ok. So, note that gamma of 0 is equal to A and gamma of 1 is equal to identity. So, let p of t p equal to determinant of gamma of t ok. So, then this is a polynomial. So, this is a non-zero polynomial with coefficients complex numbers right and this polynomial is non-zero because non-zero as p of 0 is equal to determinant of gamma of ok. Let us look at 1 or even gamma of 0 is fine, but gamma of 0 we know is A which is not equal to 0 because A is in G L and C right. So, thus p of t has at most finitely many roots has finitely many roots has sorry has at most n roots right since degree of p of t is less than equal to n. So, if you make the complex plane then here we have 0 and here we have 1 and let us say the roots of p r lambda 1 up to some lambda r where r is less than equal to n right. So, this is lambda 1 lambda 2 lambda 3 up to lambda r ok and none of the lambda i's can be 0 or 1 because at p of 0 is determinant of A which is non-zero and p of 1 is determinant of identity is non-zero ok. So, we can simply take a path we can take a path joining 0 and 1 which misses all these lambda i's yeah in the complex plane. So, let us. So, let S from 0 1 to C be a continuous path such that S of 0 is equal to 0, S of 1 is equal to 1 and the gamma i's do not belong to the image of S ok. So, then gamma compose S this from 0 1 to m and C the image lands in G L and C because if for any t as 0 1 the determinant of gamma of S of t is non-zero as determinant of gamma of S of t is equal to p of S of t which is non-zero right p of p of a complex number p lambda is equal to 0 if and only if lambda is in the set lambda 1 up to lambda r and the image of S misses all these lambda i's. So, therefore, the determinant is never 0 yeah thus gamma compose S from 0 1 the image lands inside G L and C and clearly gamma compose S of 0 is equal to gamma of 0 which is equal to A and gamma compose S of 1 is equal to gamma of 1 is equal to identity right. So, thus every matrix A in G L and C can be joined to identity by a continuous path in G L and C. So, thus G L and C is path connected. So, therefore, using this nice trick we saw that G L and C is path connected and this much easier to prove than the case of G L and R plus ok. So, let us make this simple observation. So, this completes a proof let X be path connected let F from X to Y be a continuous map. So, then F of X contained in Y with the subspace topology. So, let us see this this is easy to prove. So, recall that we had seen and we have used it many times now if we view F as a map from X to F of X where F of X has a subspace topology from Y then F is continuous. So, now we want to show that F of X is path connected. So, if we choose. So, let F of X 1 and F of X 2 be any 2 points in F of X right. So, we have X over here and here we have X 1 and X 2 and here we have let us say F of X this F of X 1 and this F of X 2 right. So, since X is path connected there exists a path which joins X 2 X 1 and this F since X is path connected there exist gamma from 0 1 to X says that gamma of 0 is X 1 and gamma of 1 is X 2 right. So, this implies that F compose gamma from 0 1 to F of X is a continuous path joining F of X 1 and F of X 2. So, this implies that F of X is path connected ok. So, this completes the proof. So, as a corollary of this we have if X is path connected and F from X to Y is a surjective continuous path then Y is path connected right. This corollary is immediate because since F is surjective it follows that Y is equal to F of X and now we just use the previous lemma. So, as a corollary of this corollary we have these nice applications SL and R these n cross n matrices says that determinant of A is equal to 1 is path connected. So, in order to show that SL and R is path connected all that we have to do is we have to construct using the previous corollary a surjective map from a path connected space to SL and R. So, we will simply take this map to SL and R. So, this map is. So, if I take a matrix A yeah A gets map 2. So, this is A. So, we just divide all the entries in the first column with determinant of A and the other entries are as it is. So, let us call this map F yeah. So, clearly the image of F ok. So, determinant of F of A is equal to since the first column has been scaled by one upon determinant of A there is equal to one upon determinant of A into determinant of A which is 1 ok. Now SL and R is contained inside GL and R and here we have F GL and R plus sorry SL and R right and this composition is equal to identity right. So, this implies that F is surjective. So, it only remains to show that F is continuous, but that is easy to see because. So, SL and R has a subspace topology from M and R SL and R is a close subspace of M and R and it has a subspace topology and therefore, to show that F is continuous it is enough to show that continuous if and only if I compose F is continuous and to say that I compose F is continuous it is enough to show that the coordinate functions are continuous yeah, but so to show I compose F is continuous enough to show that the coordinate functions are continuous yeah, but what are the coordinate functions. So, if we start with A then the coordinate functions of I compose F of A. So, the first in the first column the entries are of this type A I 1 divided by determinant of A, but one upon determinant of A is once again a continuous function. So, since determinant does not vanish on zero naught plus is continuous and not 0 this implies the function A goes to one upon determinant of A is continuous right. So, this implies. So, if I take A and I send it to A I 1 divided by determinant of A this continuous right and similarly A goes to A I j where j is strictly greater than 1 is also continuous. So, the F thus the coordinate functions of I compose F are continuous this implies I compose F is continuous this implies F is continuous yeah, and since g l and r plus is path connected and F is subjective and continuous this implies s l and r is. So, this is one application. So, similarly we can also show that s l and c this n cross n matrix with complex entries such that determinant of A is equal to 1 is path connected right. Again we use the same map from g l and c to s l and c where we just scale the first column by one upon determinant of A. Okay. So, we will end this lecture here and in the next lecture we will talk about compactness.