7. Monday 2/24: Van Kampen s Theorem The Proof

Transcription

1 7. Monday 2/24: Van Kampen s Theorem The Proof 25 Recall the statement of Van Kampen s Theorem. Let p X, and let {A α : α A } be a cover of X by path-connected open sets such that p A α for every α. We have a commutative diagram of groups, which looks in part like this (where the i s and j s are the group homomorphisms induced by inclusions of spaces). (7.1) π 1 (A α A β ) i αβ π 1 (A α ) F = α π 1 (A α ) i βα π 1 (A β ) j α Φ j β Van Kampen s Theorem: π 1 (X) π 1 ( ) = π 1 (, p) (1) If every pairwise intersection A α A β is path-connected, then the map Φ is surjective. (2) If in addition every triple intersection A α A β A γ is path-connected, then ker Φ = N := iαβ [f] i βα [ f] : α, β A and so π 1 (X) = F/N.

2 26 Proof of (1). Let f : I X be a loop based at p. Every s I has a neighborhood mapped by f into some U α. By compactness of I, there exist numbers 0 = s 0 < s 1 < < s m = 1 and indices α 1,..., α m such that f([s i 1, s i ]) A αi i [m]. Let f i = f [si 1,s i ], so that f = f 1 f 2 f m. For each i [m], the set A i A i+1 is path-connected, hence contains a path g i from p to f(s i ). Therefore f = f 1 f 2 f m = (f 1 g 1 ) (g 1 f 2 g 2 ) (g m 2 f m 1 g m 1 ) (g m f m ) π 1 (A 1, p) π 1 (A 2, p) π 1 (A m, p) im Φ. A 2 f 2 g 1 A 3 p f 1 g 2 A 1

3 Proof of (2). Let [f] π 1 (X). Say that a factorization of [f] is an expression [f 1 ] [f 2 ] [f n ] that maps to [f] via Φ. Here I am using to denote concatenation of letters to make a word in α π 1 (A α ). That is, each [f i ] belongs to some π 1 (A α ), and f f 1 f 2 f n. We want to show that any two factorizations of [f] are related by operations of the following forms: Type A : If f i : I A α A β, then we can regard the letter [f i ] as coming either from π 1 (A α ) or from π 1 (A β ). This amounts to inserting an element of N into f, namely i αβ [f i ] i βα [f i ]. Type B : If two consecutive letters in the factorization come from the same A α, we can multiply them. This, of course, doesn t change the element of F we re talking about. 27 So, suppose we have two factorizations [f] = Φ ([f 1 ] [f k ]) = Φ ([f 1] [f l]). In particular, there is a path-homotopy of p-loops H : I I X, h t (s) = H(s, t), such that h 0 = f 1 f k and h 1 = f 1 f l. Schematically, here s what this looks like: f 1 f 2 f t I x I f f 1 2 f k The dots on the top and bottom lines are the breakpoints between successive f i s or f i s.

4 28 Now, we do something clever. Partition I I into a finite grid of finitely many little rectangles R i such that (7.2) R i : i A : H(R i ) A i. (By continuity of H, we can put such a rectangle around each point in I I, then choose a finite subcover, then subdivide if necessary.) Subdivide more by adding vertical lines at all the breakpoints, and at least two horizontal lines. f f 1 2 f t f f 1 2 f k Now, we do something exceedingly clever. For all of the vertical lines not in the first or last row, give them a little nudge to one side so they don t match up. We can do this while still retaining the condition (7.2). Number the rectangles R 1,..., R mn as shown, where m is the number of columns and n is the number of rows. f f 1 2 f nm m+1 2m 1 2 f f 1 2 k f m

5 Let γ k be the path from (0, 0) to (1, 1) along the cell walls that separates rectangles R 1,..., R k from R k+1,..., R mn. (For example, the thick red path shown in the figure above is R m+1.) Thus H γ k is a closed path in X with basepoint p, and all the paths H γ k are path-homotopic. Each γ k can be written as γ k = e 1 e 2 e N where each e i is the path in X given by part of a side of one rectangle, say from v i 1 to v i. For each v i, choose some path g i in X from p to F (v i ). Each v i belongs to at most three rectangles, so we can require g i to stay in the intersection of the corresponding three A s. (Wasn t that clever of us?) Then each γ k can be factored as γ k = e 1 e N = Φ(e 1 e N ) ( ) = Φ [e 1 g 1 ] [g 1 e 2 g 2 ] [g N 2 e N 1 g N 1 ] [g N 1 e N ] Recall that means concatenation of letters in the free product F, while means concatenation within one of its free factors. To pass from the factorization for γ k to that of γ k+1, we have to trade the south and west sides of R k+1 for the north and east sides. We can do this by regarding the letters in the south and west sides as now coming from π 1 (A k+1 ) instead of wherever they came from in the factorization of γ k (this is a type-a move); using the group structure of π 1 (A k+1 ) to trade the letters in the south and west sides for the north and east ones (this is a type-b move). 29

6 30 R k+1 Now let s look at the path γ 0, which consists of the bottom and right edges of I I. The right edge is a stationary path, so forget about it. For each vertex v i on the bottom edge of I I, we have so far only required g i to lie in two of the A s. Let s also require it to lie in the same one whose fundamental group contains the letter f i (which came from the factorization of f given in advance). That says that For example, if f 3 = e 1 e 2 e 3, then the factorization begins [e 1 g 1 ] [g 1 e 2 g 2 ] [g 2 e 3 g 3 ] where g 1 is a path in A 1 A 2 and g 2 is a path in A 2 A 3. But in fact we can require g 1 and g 2 to be paths in A 1 A 2 A α and A 2 A 3 A α, where π 1 (A α ) is the group containing the letter f 3. We also may as well assume that g 3 is the stationary path. So the partial factorization shown above can be replaced (with type-a moves) with one in π 1 (A α ), and then simplified to the single letter [e 1 e 2 e 3 ] = [f 1 ] π 1 (A α ). More generally, if v i is a breakpoint then we take g i to be the constant path, and if v i is not a breakpoint then we require g i to lie in A α for whichever π 1 (A α ) contains the letter f j to which the edges at v i contribute. Then parenthesizing the factorization of f at the breakpoints shows that it is equivalent to [f 1 ] [f k ]. Playing the same game at the top of the square shows that the factorization of γ nm is equivalent to [f 1] [f l ].