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Right) Modules Over a Ring R; in Particular He Defined Chain Maps As the Morphisms in This Category

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Right) Modules Over a Ring R; in Particular He Defined Chain Maps As the Morphisms in This Category FUNDAMENTAL THEOREMS ON CHAIN COMPLEXES SAMUEL WUTHRICH¨ 1. Introduction Last week Andy Baker introduced the category of chain complexes of (right) modules over a ring R; in particular he defined chain maps as the morphisms in this category. The first section of today’s talk gives an important criterion for components of a chain map between two chain complexes to be isomorphisms. In the second section we will see how a short exact sequence of chain complexes gives rise to a long exact sequence linking the homology groups. If we speak of chain complexes, in particular of (short) exact sequences, we will always mean chain complexes of right modules over an arbitrary ring R. 2. Comparing exact sequences We begin with a few examples to get familiar with chain maps between short exact sequences. In all the examples R is Z. Assume we are given two short exact sequences and a homomorphism between the two middle terms which we would like to extend to a chain map: / i / p / / 0 A B C 0 (2.1) a b c p / i0 / 0 / / 0 A0 B0 C0 0. We find that this is not always possible, consider for example the situation / 3 / / Z / 0 Z Z / 3 0 4 0 / Z / Z / Z/4 / 0. We certainly need to know that b(i(A)) i0(A0). An easy check shows that this condition is exactly what we need: It implies the existence⊂ of a unique chain map extending our given homomorphism. Put another way, given maps a and b such that the left square in (2.1) commutes, there is a unique “fill-in map” c, and given compatible maps b and c there is a unique “fill-in” a. However it is not true that given a and c imply the existence of a matching b in the middle; look for instance at i p 0 / Z / Z Z/2 / Z/2 / 0 ⊕ (2.2) 2 0 / Z / Z / Z/2 / 0 where i is the inclusion of the first summand and p is the projection onto the second. Indeed, the element (0, 1)¯ Z Z/2 should map to an odd integer 2n + 1 to make the right square commutative. But∈ this⊕ isn’t allowed, because 0 = (0, 2)¯ has to go to zero. This is not just bad chance; for a situation as in (2.2) we have the following fact: Date: 22/10/02. 1 2 SAMUEL WUTHRICH¨ Proposition 1 (Short Five Lemma). Assume we are given a chain map between two short exact sequences of R-modules i p 0 / A / B / C / 0 a b c p / i0 / 0 / / 0 A0 B0 C0 0. If a and c are isomorphisms then so is b. Proof. We begin by showing that b is a monomorphism. We have ker(b) ker(p0b) = ker(cp) = ker(p) = im(i), ⊂ so the restriction of i to ker(bi) induces an isomorphism ker(bi) ∼= ker(b) im(i) = ker(b). −→ ∩ We conclude ker(b) ∼= ker(bi) = ker(i0a) = 0, hence b is monomorphic. To show that b is epimorphic, we first note that the composition p0b = cp is epi. Hence p0 induces an isomorphism = im(b) + ker(p0) / ker(p0) ∼ C0, −→ which shows that im(b) + ker(p ) = B . But im(b) im(bi) = im(i a) = im(i ) = ker(p ), so 0 0 ⊃ 0 0 0 B0 = im(b). £ Note however that it may happen that for given isomorphisms a and c there is still no fill-in b, even if the two middle-terms are isomorphic! An example is the following: 3 0 / Z / Z / Z/3 / 0 / 3 / / / 0 Z − Z Z/3 0. Now in general we don’t have just short exact sequences but exact sequences of arbitrary length. For these we have Proposition 2 (Five Lemma). Assume we have a commutative diagram of R-modules a1 / a2 / a3 / a4 / A1 A2 A3 A4 A5 f1 f2 f3 f4 f5 b1 / b2 / b3 / b4 / B1 B2 B3 B4 B5. with exact rows. If f1, f2, f4 and f5 are isomorphisms then so is f3. Proof. The given diagram gives rise to two short exact sequences, linked by a chain map: / a¯2 / a3 / / 0 A2/ ker(a2) ∼= coker(a1) A3 im(a3) = ker(a4) 0 ¯ f2 f3 f˜4 ¯ / b2 / b3 / / 0 B2/ ker(b2) ∼= coker(b1) B3 im(b3) = ker(b4) 0. (The maps labeled f¯2 and f˜4 exist because f2(ker(a2)) ker(b2) and f4(ker(a4)) ker(b4).) In ⊂ ⊂ fact, f2 maps ker(a2) isomorphically onto ker(b2), because f2 ker(a2) = im(a1) im(f2a1) = im(b1f1) = im(b1) = ker(b2); −→∼= similarly for f4. Hence f¯2 and f˜4 are isomorphisms, so the Short Five Lemma implies that f3 is an isomorphism as well. £ FUNDAMENTAL THEOREMS ON CHAIN COMPLEXES 3 3. The long exact homology sequence Last week we saw in an example that the homology functor does not preserve short exact sequences, i.e. if i p 0 K0 K K00 0 → −→ −→ → is a short exact sequence of chain complexes, the induced sequence in homology i p 0 H(K0) ∗ H(K) ∗ H(K00) 0 → −→ −→ → is not short exact in general. To study the phenomenon, let us first restrict attention to the case of chain complexes with just two non-trivial entries. So assume we are given a commutative diagram of the form 0 0 0 / i1 / p1 / / 0 K10 K1 K100 0 (3.1) d10 d1 d100 / i0 / p0 / / 0 K0 K0 K000 0 0 0 0, where the rows are exact. In this case, homology is given by H1(K0) ∼= ker(d10 ) and H0(K0) ∼= coker(d10 ), analogously for K and K00. This allows us to extend the given diagram to one which includes the homology modules and whose columns are exact, namely 0 0 0 (i1) (p1) / ∗ / ∗ / / 0 ker(d10 ) ker(d1) ker(d100) 0 / i1 / p1 / / 0 K10 K1 K100 0 (3.2) d10 d1 d100 / i0 / p0 / / 0 K0 K0 K000 0 (i0) (p0) / ∗ / ∗ / / 0 coker(d10 ) coker(d1) coker(d100) 0 0 0 0 In the example we considered last week, the map (p1) was not epimorphic. What can we do with this information? ∗ Assume we have an element x00 ker(d100) which has no pre-image under (p1) . In other words, 1 ∈ 1 ∗ p− (x ) ker(d1) = . So if we pick any element x p− (x ), we get a non-trivial element 1 00 ∩ ∅ ∈ 1 00 d1(x) K0. Because ∈ p0(d1(x)) = d100(p1(x)) = d100(x00) = 0, there is a (non-trivial) y K such that i0(y ) = d1(x). 0 ∈ 0 0 Now the whole point is that the residue class of y0 in coker(d10 ), which is still non-trivial (check this!) and therefore interesting, is independent of the choice of a pre-image x of x00: Assume 1 1 1 x1, x2 p− (x ) are two arbitrary preimages of x . Set y1 = i− (d1(x1)) and y2 = i− (d1(x2)); ∈ 1 00 00 0 0 4 SAMUEL WUTHRICH¨ we need to show that y1 y2 im(d ). Because p1(x1 x2) = p1(x1) p1(x2) = x x = 0, − ∈ 10 − − − there is a ∆ K with i1(∆) = x1 x2. Now ∈ 10 − i0(d0 (∆)) = d1(i1(∆)) = d1(x1 x2) = d1(x1) d1(x2) = i0(y1 y2), 1 − − − so injectivity of i0 implies d10 (∆) = y1 y2, so indeed y1 y2 im(d10 ). To summarize, we have shown that− the assignment − ∈ 1 p1(x) [ i− (d1(x))] 7→ 0 is a well-defined map ∂1 : ker(d00) coker(d0 ), 1 → 1 indeed a homomorphism. According to the next proposition, we can explain the deviation of exactness in homology in terms of ∂1. Proposition 3 (Snake Lemma). Assume we are given a commutative diagram of the form (3.1). Then the following sequence is exact: (i1) (p1) ∗ / ∗ / 0 ker(d10 ) ker(d1) ker(d100) → LLL LL∂ LL1L LL% (i0) / (p0)/ coker(d ) ∗ coker(d1) ∗coker(d ) 0 10 100 → We will not prove the Snake Lemma on its own, because it is a special case of a more general result, which considers chain complexes of arbitrary length: Theorem 1 (Long exact homology sequence). Assume i p 0 (K0, d0) (K, d) (K00, d00) 0 → −→ −→ → is a short exact sequence of chain complexes. (1) For every n, there is a connecting homomorphism ∂n : Hn(K00) Hn 1(K0), → − defined by 1 [pn(x)] [ in− 1(dn(x))]. 7→ − (2) The sequence (in) (pn) ∂n ... Hn(K0) ∗ Hn(K) ∗ Hn(K00) Hn 1(K0) ... → −−−→ −−−→ −→ − → is exact. It is called the long exact homology sequence. (3) This construction is natural, i.e. if / i / p / / 0 K0 K K00 0 f 0 f f 00 / j / q / / 0 L0 L L00 0 is a commutative diagram of chain complexes with exact rows, then the diagram K (in) (pn) ∂n / ∗ / ∗ / / / ... Hn(K0) Hn(K) Hn(K00) Hn 1(K0) ... − (fn0 ) (fn) (fn00) (fn0 1) ∗ ∗ ∗ − ∗ L (jn) (qn) ∂n / ∗ / ∗ / / / ... Hn(L0) Hn(L) Hn(L00) Hn 1(L0) ... − K L commutes too; here ∂n and ∂n are the connecting homomorphism which are associated according to (1) to the short exact sequences 0 K0 K K00 and 0 L0 L L 0 respectively. → → → → → → 00 → FUNDAMENTAL THEOREMS ON CHAIN COMPLEXES 5 Proof. (1) Exactly as we did for the Snake Lemma, we construct a homomorphism ˜ ∂n : ker(dn00 ) coker(dn0 1) → − ˜ 1 for each n, by setting ∂n(pn(x)) = [ in− 1(dn(x))]. − 1 ˜ First we note that an element y0 = in− 1dn(x) representing the residue class of ∂n(pn(x)) is a − cycle, i.e.
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