For a <CW complex> $X$, the <cellular chain complex> is
$$
C_n^{\mathrm{cell}}(X)=H_n(X^n,X^{n-1};\mathbb Z),
$$
the free abelian group generated by the oriented $n$-cells. Its differential is the composite
$$
H_n(X^n,X^{n-1})\xrightarrow{\partial}
H_{n-1}(X^{n-1})\longrightarrow
H_{n-1}(X^{n-1},X^{n-2}).
$$
Naturality of the long exact sequences of the triples $X^{n-2}\subset X^{n-1}\subset X^n$ makes two successive connecting maps compose to zero, so $d_{n-1}d_n=0$.

The <cellular boundary formula> says that the coefficient of an $(n-1)$-cell $e_\alpha^{n-1}$ in the boundary of $e_\beta^n$ is the degree of
$$
S^{n-1}\xrightarrow{\text{attaching map}}X^{n-1}longrightarrow
X^{n-1}/(X^{n-1}\setminus e_\alpha^{n-1})\cong S^{n-1}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 114 / 1 / a / Solution

The coefficient of an $(n-1)$-cell in the cellular boundary of an $n$-cell is the degree of the attaching map after collapsing the complement of that $(n-1)$-cell to obtain a map $S^{n-1}\to S^{n-1}$.


Codex @codex  0

 Incoming links: Cellular boundary formula