Homology, Long Exact Sequence

Long Exact Sequence

Let C D and E be chain complexes, with chain maps f from C to D, and g from D to E. In fact, the chain maps define a short exact sequence, which is short exact at every level.

0 → C → D → E → 0

0 → C_i → D_i → E_i → 0 { at every level }

Let Ck_i be the kernel of C_i mapping into C_i+1. Let Ca_i be the image of C_i-1 from above. Thus the homology group on C_i is Ck_i/Ca_i.

Establish similar notation for D and E. Thus the homology group on E_i is Ek_i/Ea_i.

Remember that the chain maps, from C to D to E, induce corresponding homomorphisms on the homology groups. Build a new table with homology groups running down the columns, and induced homomorphisms running across. Thus the 7^th row looks like this.

Ck₇/Ca₇ → Dk₇/Da₇ → Ek₇/Ea₇

When this table is traversed in row order, an exact sequence of homology groups emerges. This is called the exact homology sequence or the long exact sequence. The proof is tedious, and involves lots of diagram chasing.

I'm going to prove the theorem for row 7, which is my favorite row. Remember that row 6 is above and row 8 is below.

Apply the serpent lemma, and Ck₇ → Dk₇ → Ek₇ is exact. Restrict this row to these kernels. This causes no trouble from above, since row 6 maps into kernels. Furthermore, the chain maps f and g carry kernels to kernels. Thus, temporarily, let row 7 look like this.

0 → Ck₇ → Dk₇ → Ek₇ → ?

Apply the serpent lemma to rows 6 and 7, and Ck₇/Ca₇ → Dk₇/Da₇ → Ek₇/Ea₇ is exact. That takes care of the central homology group.

The serpent lemma generates a function from Ek₇ to C₈/Ca₈. Call this function h. Let x represent an element in the range of h, a coset of Ca₈ in C₈. The serpent naturally carries x to something in D₈, which comes from D₇. That's how h was constructed. So moving down to D₉ yields 0. Therefore x becomes 0 in C₉. In other words, x lies in Ck₈. The image of h lies in Ck₈, and is part of the homology group Ck₈/Ca₈.

What does h do to Ea₇? Given y in Ea₇, back up to find x in D₇. This is how h works, and any x in the preimage of y will do. This includes the x that appears by pulling up to E₆, back to D₆, and down to D₇. Now the image of x in D₈ is 0, which pulls back to 0 in C₈. Therefore h maps Ea₇ into the 0 coset of Ca₈, which is Ca₈. We have a valid map from the homology of E₇ into the homology of C₈.

The serpent lemma tells us the elements of Ek₇ are exact. Some y in Ek₇ is in the image of Dk₇ iff it is in the kernel of h. The same is true when y represents a coset of Ea₇. Thus the homology group on the right is exact.

Let j be the function that follows h, mapping the homology of C₈ into the homology of D₈. Restrict the 8^th row to its kernels, and j is the function that is implied by the serpent lemma. The lemma also tells us x is in the image of h iff it is in the kernel of j. View h as a function of homology, rather than a simple function on Ek₈, and the homology group on C₈ becomes exact.

The induced homology homomorphisms are exact ad D₇, E₇, and C₈. This holds for every row, building a long exact sequence that runs forever in both directions.

Long Exact Split

Continue the above, but this time assume 0 → C → D → E → 0 is split exact. The group D₇ is the direct product of the image of C₇ and the quotient group, which happens to be isomorphic to E₇. Restrict D to its subchains, the image of C and the preimage of E, and the homology groups are restricted accordingly. Each homology group is the direct product of its restricted subgroups.

We're using the fact that f and g are chain maps. Going from C₆ to C₇ to C₈ is mirrored by the homomorphisms on their images in D₆, D₇, and D₈. Thus the homology group on C₇ is isomorphic to the first component of the homology group of D₇. The same holds for the reverse maps from E back to D. Therefore the homology of the central column is the direct product of the homologies on either side. In other words, the long exact sequence is split exact on the middle column.