Composite of consecutive maps of cellular chain complex is zero

Statement

Statement independent of the language of topological spaces

Suppose $A\subseteq B\subseteq C\subseteq D$ are topological spaces, each having the subspace topology from the next. Suppose $n$ is a (nonnegative, though it doesn't matter) integer. Consider the following three cases of the long exact sequence of homology of a pair (we write different parts of the long exact sequences because we need different parts for the later steps):

• The pair $(D,C)$:

$\ldots \to H_{n+1}(C)\to H_{n+1}(D)\to H_{n+1}(D,C){\stackrel {p_{1}}{\to }}H_{n}(C)\to H_{n}(D)\to H_{n}(D,C)\to \ldots$

• The pair $(C,B)$:

$\ldots \to H_{n+1}(C,B)\to H_{n}(B)\to H_{n}(C){\stackrel {p_{2}}{\to }}H_{n}(C,B){\stackrel {p_{3}}{\to }}H_{n-1}(B)\to H_{n-1}(C)\to H_{n-1}(C,B)\to H_{n-2}(B)\to \ldots$

• The pair $(B,A)$:

$\ldots \to H_{n}(B,A)\to H_{n-1}(A)\to H_{n-1}(B){\stackrel {p_{4}}{\to }}H_{n-1}(A,B)\to H_{n-2}(A)\to \ldots$

Define:

• $f:=p_{2}\circ p_{1}$
• $g:=p_{4}\circ p_{3}$

Then $g\circ f=0$.

Statement in terms of cellular homology

This is a special case of the previous statement, where $A=X^{n-2},B=X^{n-1},C=X^{n},D=X^{n+1}$ are the successive skeleta of a cellular filtration of a topological space $X$. In this case, the map $f$ defined above is the boundary map $\partial _{n+1}$ and the map $g$ defined above is the boundary map $\partial _{n}$ in the Cellular chain complex (?). The statement is thus that $\partial _{n+1}\circ \partial _{n}=0$, i.e., the cellular chain complex is indeed a chain complex.

Proof

The proof is a one-liner: $g\circ f=(p_{4}\circ p_{3})\circ (p_{2}\circ p_{1})=p_{4}\circ (p_{3}\circ p_{2})\circ p_{1}$. The intermediate composite $p_{3}\circ p_{2}$ is zero since it is a composite of successive maps in a long exact sequence of homology. Hence, the overall composite is the zero map.