Incidence number of two CW cells

For $n\ge 2$ and oriented cells $e_{\alpha }^n$ and $e_{\beta }^{n-1}$, collapse the complement of $e_{\beta }^{n-1}$ in $X^{n-1}$ and compose the attaching map of $e_{\alpha }^n$ with the resulting quotient to $S^{n-1}$. Its induced endomorphism of oriented ${\tilde{H}}_{n-1}(S^{n-1};\mathbb{Z})$ is multiplication by a unique integer, denoted $[e_{\alpha }^n:e_{\beta }^{n-1}]$.

For $n=1$, choose the characteristic parametrization so the interval from $-1$ to $+1$ represents the chosen generator of $e_{\alpha }^1$. Write the chosen zero-cell generator as $e_v={\epsilon }_v[v]$, with ${\epsilon }_v\in \{1,-1\}$ and $[v]$ the canonical point class in $H_0(X^0;\mathbb{Z})$. For $v=e_{\beta }^0$ set $[e_{\alpha }^1:v]={\epsilon }_v({\mathbf{1}}_{\{{\chi }_{\alpha }(+1)=v\}}-{\mathbf{1}}_{\{{\chi }_{\alpha }(-1)=v\}})\mathrm{.}$ Indeed the singular boundary is $[{\chi }_{\alpha }(+1)]-[{\chi }_{\alpha }(-1)]$ and $[v]={\epsilon }_v{e}_v$, so this is precisely its coefficient in the chosen cellular basis. Canonical positive vertex generators give the usual terminal-minus-initial formula. Equal endpoints cancel for either sign. The negative vertex choice is an algebraic basis change, not a based self-map of $S^0$ of degree $-1$.