FURTHER REMARKS ON SYSTEMS OF INTERLOCKING EXACT SEQUENCES

In a system of interlocking sequences, the assumption that three out of the four sequences are exact does not guarantee the exactness of the fourth. In 1967, Hilton proved that, with the additional condition that it is differential at the crossing points, the fourth sequence is also exact. In this paper, we trace such a diagram and analyze the relation between the kernels and the images, in the case that the fourth sequence is not necessarily exact. Regarding the exactness of the fourth sequence, we remark that the exactness of the other three sequences does guarantee the exactness of the fourth at noncrossing points. As to a crossing point p, we need the extra criterion that the fourth sequence is differential. One notices that the condition, for the fourth sequence, that kernel ⊇ image at p turns out to be equivalent to the “opposite” condition kernel ⊆ image. Next, for the kernel and the image at p of the fourth sequence, even though they may not coincide, they are not far different—they always have the same cardinality as sets, and become isomorphic after taking quotients by a subgroup which is common to both. We demonstrate these phenomena with an example.

Note that one derives Theorem 1.1 through standard diagram chasing. However, the criterion that the fourth sequence is differential at its crossing points is not needed when showing the exactness at non-crossing points; thus, Proposition 1.2. Suppose given 4 sequences of which 3 are long exact, forming a commutative diagram I  I  I  I  I  I  I  I  I  I   I  I  I  I  I  I  I  I  I   I  I  I  I  I then the fourth is automatically exact at all points other than the crossing points p.
Proof. In the commutative diagram of sequences, thus, T  T  T  T  T  T  T  T  T  T  T  T   T  T  T  T  T  T  T  T  T  T  T  T   T  T  T  T  T  T  T  T  T  T  T assume, without real loss of generality, that the α-, β-, and γ-sequences of the ∂-sequence are exact.
To show the exactness of the ∂-sequence at A i , first one sees that As to the exactness of the ∂-sequence at C i+1 , first it is differential because Hereafter, it remains to examine the exactness of the ∂-sequence at the crossing points p, in diagram (1.2). Theorem 1.1 says that, with the assumption that the α-, β-, and γsequences are exact, the fourth sequence, the ∂-sequence, is also exact if it is differential at p. We trace the diagram and make the following expansion: In the commutative diagram of sequences, thus, T  T  T  T  T  T  T  T  T  T  T  T   T  T  T  T  T  T  T  T  T  T  T  T   T  T  T  T  T  T  T  T  T  T  T assume that the α-, β-, and the γ-sequences are exact. Then image ∂ i ⊆ ker ∂ i+1 if and only if ker ∂ i+1 ⊆ image ∂ i . Thus, the following are equivalent: (i) the ∂-sequence is exact at the crossing points p, Before proving Theorem 1.3, we note that it is easy to verify that the fourth sequence in diagram (1.1) is differential at the crossing points; hence the sequence is exact: For the composite homomorphism π n (A, B) → π n (X, B) → π n (X, A) is induced by the composite map which coincides with the composite map Proof of Theorem 1.3. First we assume that the fourth sequence is differential at B i and show By the assumption that ker ∂ i+1 ⊆ image ∂ i , we find an a ∈ A i such that Thus, there is a c ∈ C i−1 such that due to the facts that c − β i−1 (b) ∈ ker γ i and the γ-sequence is exact. Finally, the calculation that completes the proof that the ∂-sequence is differential at the crossing points.
Now we have shown that image ∂ i ⊆ ker ∂ i+1 if and only if ker ∂ i+1 ⊆ image ∂ i , the assertion that the ∂-sequence is exact at the crossing points p, it is differential at p, and ker Hence, in a system of interlocking sequences, (1.2), the fact that three out of the four sequences are exact does not guarantee the exactness of the fourth at the crossing points p.
One needs the condition that the sequence is differential at p to assure its exactness. The next couple of examples demonstrate the necessity of this extra criterion. where ι 1 , ι 2 are the inclusions into the first and the second factors, respectively, 2 is the projection onto the second factor, and ∂ = 1 where ι 1 is the inclusion into the first factor, 1 , 2 are the projections onto the first and the second factors, respectively, and ∆ = {1 G , 1 G }; that is, ∆(g) = (g, g) where g ∈ G.
Note that, in either example, it is easy to verify all the necessary commutativity relations; yet, the exactness of the fourth sequence · · · / / • / / • / / · · · fails at the crossing point because none of these equivalent relations, thus, holds.

Kernels vs. Images at the Crossing Points
As discussed in Section 1, in such a commutative diagram (1.3), where the α-, β-, and γ−sequences are exact, the fourth sequence, the ∂-sequence, may not be exact at the crossing points p. Nevertheless, in Example 1.4, Example 1.5, and a few other examples, we notice that, even though image ∂ i and ker ∂ i+1 might not coincide, they seem to be isomorphic in most cases and they always have the same cardinality as sets. The following theorem assures our assertion and says the close relation between image ∂ i and ker ∂ i+1 .
Theorem 2.1. In the commutative diagram of sequences, thus, / / β i+1 P P P P P P P P P P P P P P P P P P P P P P P P To assure that χ is well defined, assume that there are elements a and a in A i such that α i (a) = γ i+1 (b) = α i (a ). Then α i (a − a ) = 0, so there exists an x ∈ B i−1 such that We next show that the map χ is monomorphic: Suppose given a , there is an Next, since β i (c) = 0 and the β-sequence is exact, there is a y ∈ B i−1 such that , and therefore, the map χ is monomorphic.
is an isomorphism, it is evident that image ∂ i and ker ∂ i+1 must have the same cardinality as sets.
It is unfortunate that, in (1.2), the fact that three out of the four sequences are exact does not promise the exactness of the fourth. Nevertheless, the only place that we need to examine is at the crossing points p. Even though the kernel and the image may not agree here, they are not far different -they always have the same cardinality as sets, and they become isomorphic after taking quotients by a subgroup which is common to both. We close the discussion with an example that demonstrates these phenomena.