I'm reading "Algebraic Topology" by Allen Aatcher and I came across the following paragraph which supposedly explains that the Mayer-Vietoris exact sequences exist also for relative homology:

$"$If one has a pair of spaces $(X, Y ) = (A ∪ B, C ∪ D)$ with $C ⊂ A$ and $D ⊂ B$ , such that $X$ is the union of the interiors of $A$ and $B$ , and $Y$ is the union of the interiors of $C$ and $D$, then there is a relative Mayer–Vietoris sequence

$··· → H_n(A ∩ B, C ∩ D) _Φ →H_n(A, C) ⊕ Hn(B, D) _Ψ →Hn(X, Y ) _∂ →···$

To derive this, consider the commutative diagram enter image description here where $C_n(A + B, C + D)$ is the quotient of the subgroup $C_n(A + B) ⊂ C_n(X)$ by its subgroup $C_n(C + D) ⊂ C_n(Y)$ .Thus the three columns of the diagram are exact.

We have seen that the first two rows are exact, and we claim that the third row is exact also, with the maps $ϕ$ and $ψ$ induced from the $ϕ$ and $ψ$ in the second row.

Since $ψϕ = 0$ in the second row, this holds also in the third row, so the third row is at least a chain complex. Viewing the three rows as chain complexes, the diagram then represents a short exact sequence of chain complexes.

The associated long exact sequence of homology groups has two out of every three terms zero since the first two rows of the diagram are exact.

Hence the remaining homology groups are zero and the third row is exact. The third column maps to $0→C_n(Y)→C_n(X)→C_n(X,Y)→0$, inducing maps of homology groups that are isomorphisms for the $X$ and $Y$ terms as we have seen above. So by the five-lemma the maps $C_n(A+B,C+D)→C_n(X,Y)$ also induce isomorphisms on homology. The relative Mayer–Vietoris sequence is then the long exact sequence of homology groups associated to the short exact sequence of chain complexes given by the third row of the diagram.$"$

I don't understand the part in the quote box:

1) "two out of every three terms" - which terms ?

2) Why does having the two first rows exact makes the third exct? (how he used the properties of commutative diagrams here?)

Thanks in advance


migrated from mathoverflow.net May 4 '18 at 14:11

This question came from our site for professional mathematicians.


Let's denote the three rows (viewed as chain complexes) by $E,F,G$, so we have a short exact sequence of chain complexes $0 \to E \to F\to G \to 0$. So the long exact sequence on homology will look like $$ \dots \to H_n(E) \to H_n(F) \to H_n(G) \to H_{n-1}(E) \to \dots$$ Now since $E$ and $F$ are exact chain complexes, all their homology groups are zero, so all the $H_i(E)$ and $H_i(F)$ terms are zero (this is the "two out of every three terms"). So all the $H_i(G)$ terms must be zero as well.

Thus $G$ is an exact chain complex, so the third row (which is $G$) is exact. This last step doesn't use the properties of the commutative diagram, rather it uses the fact that a chain complex is exact if and only if all its homology groups are zero.


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