( 1) i [ d i ]. The claim is that this defines a chain complex. The signs have been inserted into the definition to make this work out.

Transcription

1 Mon, Apr. 2 We wsh to specfy a n : C n ()! C n (). Snce C n () s a free abelan group, the n s completely specfed by ts value on each generator, namely each n-smplex. There are n + standard nclusons d : n,! n, gven by nsertng 0 n poston n n. Defnton 5.2. The sngular boundary n : C n ()! C n () s defned n ( ) ( ) [ d ]. 0 Example 5.3. () If s a -smplex (from v 0 to v ), ( )[ d 0 ] [ d ][v ] [v 0 ]. (2) If s a 2-smplex wth vertces v 0, v, and v 2, and edges e 0, e 02, and e 2 ( )[ d 0 ] [ d ]+[ d 2 ][e 2 ] [e 02 ]+[e 0 ] The clam s that ths defnes a chan complex. The sgns have been nserted nto the defnton to make ths work out. Proposton 5.4. The boundary squares to zero, n the sense n 0. Proof. We wll use Lemma 5.5. For >j,thecomposte n 2 d j! n d! n s equal to the composte n 2 d! n d j! n. Consder the case 3, j, n 4. We have d 3 (d (t,t 2,t 3 )) d 3 (t, 0,t 2,t 3 )(t, 0,t 2, 0,t 3 )d (t,t 2, 0,t 3 )d (d 2 (t,t 2,t 3 )). Ths argument generalzes. 39

2 For the n ( n ( ) [ d ] (changng bounds) (Lemma) 0 ( n ([ d ]) 0 ( ) ( ) j [ d d j ] 0 0. j0 ( ) ( ) j [ d d j ]+ 0 j< ( ) ( ) j [ d d j ] 0 j ( ) ( ) j [ d d j ]+ ( ) ( ) j [ d d j ] j< 0 j ( ) ( ) j [ d j d ]+ ( ) ( ) j [ d d j ] j< j0 j 0 j ( ) j ( ) [ d d j ]+ ( ) ( ) j [ d d j ] 0 j We have shown that any two successve smplcal boundary homomorphsms compose to zero, so that we have a chan complex. What do we do wth a chan complex? Take homology! Defnton C n+! C n n!... s a chan complex, then we defne the nth homology group H n (C,@ )tobe H n (C,@ ):ker@ n / n+. Note that the fact n+ 0 mples that n+ s a subgroup of n, so that the defnton makes sense. A complex (C,@ ) s sad to be exact at C n f we have equalty n m@ n+. Thus the homology group H n (C,@ ) measures the falure of C to be exact at C n. Defnton 5.7. Gven a space, wedefnethesngular homology groups of to be H n (; Z) :H n (C (),@ ). Note that we only defned the groups C n () for n 0. For some purposes, t s convenent to allow chan groups C n for negatve values of n, so we declare that C n () 0 for n<0. Ths means that 0 C 0 (), so that H 0 C 0 ()/ coker(@ ). Termnology: The group n s also known as the group of n-cycles and sometmes wrtten Z n. The group m(@ n+ ) s also known as the group of boundares and sometmes wrtten B n. Wed, Apr. 4 Remark 5.8. It s worth notng that snce each C n () s free abelan and n and n+ are both subgroups, they are necessarly also free abelan. 40

3 Example 5.9. Consder. ThenC n ({ }) Z{Top( n, { })} Z for all n. The d n : C n ({ })! C n ({ }) takes the (constant) sngular n-smplex c n to the alternatng sum ( ) cn c n n even 0 n odd. In other words, the chan complex s...! Z 0! Z! Z 0! Z, so that the only nonzero homology group s H 0 ( ) Z. But already for, the chan groups are nfnte rank, and computng becomes mpractcal. On the other hand, the sngular homology groups have good propertes. For starters, we wll dscuss functoralty. Gven a map f :! Y, we can compose any sngular n-smplex of wth f to get a sngular n-smplex of Y. Ths produces a functon and therefore a homomorphsm f n :Sng n ()! Sng n (Y ) f n : C n ()! C n (Y ). It remans to dscuss how ths nteracts wth homology. Defnton 5.0. Let (C,@ C ) and (D,@ D ) be chan complexes. Then a chan map f : (C,@ C )! (D,@ D ) s a sequence of homomorphsms f n : C n! D n, for each n, such that each dagram C n f n / D n commutes for each C D n C n fn / D n Snce f n s gven by post-composton wth f, whereas each term n s gven by precomposng wth the face nclusons, t follows that the homomorphsms (f ) on the sngular chans assemble to produce a chan map. We set up ths defnton n order to get Proposton 5.. A chan map f : (C,@ C )! (D,@ D ) nduces homomorphsms f n : H n (C,@ C )! H n (D,@ D ) for each n. Proof. Let x 2 C n be a cycle, meanng C (x) 0. D (f n (x)) f n (@ C (x)) f n (0) 0, so that f n (x) sacyclend n. In order to get a well-defned map on homology, we need to show that f x s n the mage n+ C,thenf n(x) s n the mage n+ D. But f n+ C (y), then f n(x) f n (@ n+ C n+ f n+(y), whch shows that f n (x) s a boundary. There s an obvous way to compose chan maps, so that chan complexes and chan maps form a category Ch 0 (Z). Proposton 5.2. The assgnment 7! (C (),@ ) and f 7! f defnes a functor C : Top! Ch 0 (Z). 4

4 Gven the above dscusson, t only remans to show that ths constructon takes dentty morphsms to dentty morphsms and that t preserves composton. We leave ths as an exercse. Note that the sequence of homology groups H n (C,@ C ) of a chan complex s not qute a chan complex, snce there are no d erentals between the homology groups. You can thnk of ths as a degenerate case of a chan complex, n whch all d erentals are zero. But t s more common to smply call ths a graded abelan group. If and Y are graded abelan groups, then a graded map f :! Y s smply a collecton of homomorphsms f n : n! Y n. Graded maps compose n the obvous way, so that we get a category GrAb of graded abelan groups. Then Proposton 5. s the man step n provng Proposton 5.3. Homology defnes a functor H : Ch 0 (Z)! GrAb. The composton of two functors s always a functor. Thus Proposton 5.2 and Proposton 5.3 combne to yeld Proposton 5.4. Sngular homology defnes a functor H Sng : Top! GrAb. Ths mples, for nstance, that homeomorphc spaces have somorphc sngular homology groups. Fr, Apr. 6 Last tme, we dscussed how a map of spaces nduces a map on homology. Prevously, we saw that the nduced map on fundamental groups only depended on the homotopy class of the map, and we mght ask the same queston here. Proposton 5.5. Suppose that f ' g as maps homology. Corollary 5.6. If f : homology.! Y. Then f and g nduce the same map on! Y s a homotopy equvalence, then f nduces an somorphsm on Sketch of Proposton 5.5. See Theorem 3.8 of Lee for complete detals. If we have maps f,g :! Y, t would be enough to show that ther d erence f g at the level of chans always takes values n the group of boundares. Unfortunately, ths s not always true, but t turns out to be true on cycles, whch s enough to deduce the proposton. For smplcty, we consder the unversal case, n whch Y I and f and g are the nclusons at tme 0 and, respectvely. The dea s to defne a homomorphsm (called a chan-homotopy ) h n : C n ()! C n+ ( I) for all n, satsfyng the equaton h g f. If you plug n a cycle x to ths formula, you learn that g (x) f (x) s a boundary, so that f and g agree at the level of homology. When n 0, we smply take h 0 (x) to be the constand path n I from (x, 0) to (x, ). At level, f s a path n, wewshtodefneh ( ) 2 C 2 ( I) wth h 0 ( () (0)) h ( ) {} {0}. We take h ( ) to be the formal d erence of smplces wth vertces ( 0, 0), (, 0), and (, ) and ( 0, 0), ( 0, ), (, ). Smlar formulas work n hgher dmensons. 42

5 Example 5.7. We saw that the one-pont space has homology groups nonvanshng only n dmenson zero, gven by the group Z. It follows that the same s true for any contractble space, such as I n or D n or R n. 43