Algebraic Topology: the Mayer-Vietoris Sequence

Transcription

1 Algebraic Topology: the Mayer-Vietoris Sequence The Mayer-Vietoris sequence is use to fin information abuot the homology groups of a space in terms of those of its subspaces. More specifically, if K is a simplicial complex an L, M are subcomplexes of K, then we can form a long exact sequence of homology groups an homeomorphisms between them. In orer to formulate the theorem, we introuce some notation. Let K be a simplicial complex, an L, M subcomplexes of K. Let the inclusion maps i, j, k, l, the inuce chain maps an the inuce homomorphisms between homology groups be given by the following squares: L M i L C n (L M) i n C n (L) H n (L M) i H n (L) j k j n k n j k M l K C n (M) l n C n (K) H n (M) l H n (K) Theorem Mayer-Vietoris There is an exact sequence given by H n (L M) H n (L) H n (M) H n (K) H n 1 (L M) H n 1 (L) H n 1 (M) H n 1 (K) H 0 (L M) H 0 (L) H 0 (M) H 0 (K) 0 Proof: where (x) = (i (x), j (x)), (x, y) = k (x) l (y) an is constructe in the proof. We break the proof up into several steps. 1

2 Step 1: Reuce to an algebraic problem Note that for any n, the sequence is exact. To see this, note that: 0 C n (L M) αn C n (L) C n (M) βn C n (K) 0 (i) α n is injective, since i n, j n are; (ii) β n is surjective, since K = L M; (iii) ker β n = {(c, c) : c C n (L) C n (M)} = im α n. Hence we reuce to a purely algebraic result: given chain complexes C, D an E with chain maps such that the sequence α C β D E α 0 C n n Dn exact for each n, the following sequence β n E n 0 H n (C ) H n (D ) H n (E ) H n 1 (C ) H n 1 (D ) H n 1 (E ) H 0 (C ) H 0 (D ) H 0 (E ) 0 is exact. The result we seek is then a special case of this purely homological result. 2

3 In what follows, keep the following picture in min. β n+1 D n+1 E n+1 0 α n 0 C n D n E n 0 β n α n 1 β n 1 0 C n 1 D n 1 E n 1 0 α n 2 0 C n 2 D n 2 The rows are exact, an element of im is a bounary an an element of ker is a cycle. Step 2a: Construct We wish to efine a well-efine homomorphism : H n (E ) H n 1 (C ) for each n. We ll worry about whether our map is well-efine an a homomorphism later; here we just wish to allocate a member of H n 1 (C ) to each member of H n (E ). Given x H n (E ), construct (x) as follows: (i) Let y E n be a cycle representing x. (ii) Let z D n be such that β n (z) = y. (iii) Let w C n 1 be such that α n 1 (w) = (z). (iv) Let (x) = [w]. Why are we allowe to o this? Well (i) an (iv) are clearly okay. For (ii), note that z exists by surjectivity of β n. For (iii), note that β n 1 (z) = β n (z) = (y) = 0 since y is a bounary, so (z) ker β n 1 = im α n 1. Thus such a w exists, an moreover it is a cycle since 0 = 2 (z) = α n 1 (w) = α n 2 (w) an injectivity of α n 2 implies that (w) = 0. 3

4 Step 2b: is well-efine In (2a) we mae choices for y, z, w. Suppose we chosen y, z, w instea; we require [w] = [w ] for to be well-efine. Since [y] = [y ] = x, y an y iffer by a bounary, so that there is some u E n+1 with y y = (u). Then by surjectivity of β n+1 there is a v D n+1 with β n+1 (v) = u, an so we get β n (z z (v)) = y y β n (v) = y y (u) = 0 Hence z z (v) ker β n. By exactness, there is a t C n such that α n (t) = z z (v). But then α n 1 (t) = α n (t) = (z) (z ) 2 (v) = (z) (z ) = α n 1 (w w ) so by injectivity of α n 1 we have w w = (t). That is, w an w iffer by a bounary, which is precisely the statement that [w] = [w ]; so is well-efine. Step 2c: is a homomorphism Let x 1, x 2 X, an let y i, z i, w i (i = 1, 2) be the respective choices mae in the construction of (x i ). It is clear that y 1 + y 2 is a cycle representing x 1 + x 2. Since β n is a homomorphism we have β n (z 1 + z 2 ) = y 1 + y 2, an α n 1 (w 1 + w 2 ) = α n 1 (w 1 ) + α n 1 (w 2 ) = (z 1 ) + (z 2 ) = (z 1 + z 2 ) We see that (x 1 + x 2 ) = [w 1 + w 2 ] = [w 1 ] + [w 2 ] = (x 1 ) + (x 2 ), an so is a homomorphism. Step 3a: Exactness at H n (D ) The fact that im ker follows from the fact that im α n ker β n : namely, β n α n = 0 an so = 0 by functoriality of ( ). Now suppose z Z n (D ) with [z] ker. Then β n (z) B n (E ), so that β n (z) = (u) for some u E n+1. By surjectivity of β n+1 there exists v D n+1 with β n+1 (v) = u. Then β n (z (v)) = β n (z) (u) = 0, so z (v) ker β n an hence z (v) = α n (t) for some t C n. 4

5 But then α n 1 (t) = (z (v)) = 0, so that (t) = 0 by injectivity of α n 1. So t Z n (C ) an ([t]) = [z (u)] = [z], an ker im. Step 3b: Exactness at H n (E ) Let x im. Choose z Z n (D ) in the construction of (x) with ([z]) = x. Then (z) = 0, so that w = (x) = 0, an hence im ker. Conversely, if x ker then w B n 1 (C ), say w = (t) where t C n. Then (z α n (t)) = (z) α n 1 (t) = (z) α n 1 (w) = (z) (z) = 0 so z α n (t) Z n (D ); an β n (z α n (t)) = β n (z) = y since β n α n = 0. So ([z α n (t)]) = [y] = x, an so ker im. Step 3c: Exactness at H n (C ) If x H n (E ) an w, z are as efine in the construction of (x), then we have (x) = ([w]) = [(z)] = 0, so that im ker. Now suppose [w] ker H n 1 (C ). Then α n 1 (w) B n 1 (D ), say α n 1 (w) = (z), an let y = β n (z). Then (y) = β n 1 (z) = β n 1 α n 1 (w) = 0 since im α n 1 = ker β n 1. Hence y Z n (E ) an [y] satisfies ([y]) = [w]. So ker im. This proves the theorem. Acknowlegements This proof is a fleshe out version of that which I took own in my lecture notes for the Part II Algebraic Topology lecture course as given by Prof. Peter Johnstone in Michaelmas Term 2011 at the University of Cambrige. Any mistakes appearing are my own if any are foun, please sen an to clive@cnewstea.co.uk to let me know! Clive Newstea May 2012 (revise August 2012) 5