Homology and cohomology

This single identity is what makes homology well-defined: it guarantees $\operatorname{im} \partial_{n+1} \subseteq \ker \partial_n$, so the quotient in the definition of $H_n(X)$ actually makes sense as a group.

It suffices to check the claim on a single $n$-simplex $\sigma = [v_0, \dots, v_n]$ and extend by linearity. By definition, $\partial_n \sigma = \sum_{i=0}^n (-1)^i [v_0, \dots, \hat{v_i}, \dots, v_n]$, where $\hat{v_i}$ means $v_i$ is deleted.

Applying $\partial_{n-1}$ to each term and deleting a second vertex $v_j$ (with $j \ne i$) from $[v_0, \dots, \hat{v_i}, \dots, v_n]$ produces the face $[v_0, \dots, \hat{v_j}, \dots, \hat{v_i}, \dots, v_n]$ that is missing exactly the two vertices $v_i, v_j$. Carefully tracking the sign: when $j < i$ the face is deleted at position $j$ first (sign $(-1)^j$) inside a term already carrying sign $(-1)^i$, giving total sign $(-1)^{i+j}$; when $j > i$, deleting $v_j$ from the $(n-1)$-simplex $[v_0,\dots,\hat{v_i},\dots,v_n]$ removes the vertex now sitting at position $j-1$ (because $v_i$ was already removed), giving sign $(-1)^i (-1)^{j-1} = -(-1)^{i+j}$.

So $\partial_{n-1}\partial_n \sigma = \sum_{j<i} (-1)^{i+j} [\dots,\hat{v_j},\dots,\hat{v_i},\dots] + \sum_{j>i} \left(-(-1)^{i+j}\right) [\dots,\hat{v_i},\dots,\hat{v_j},\dots]$. Every face missing exactly two vertices $\{v_i, v_j\}$ with $i < j$ appears exactly twice in this double sum: once from deleting $v_i$ then $v_j$ (contributing $-(-1)^{i+j}$) and once from deleting $v_j$ then $v_i$ (contributing $+(-1)^{i+j}$), and these two contributions are exact opposites.

Every term cancels in pairs, so $\partial_{n-1}\partial_n \sigma = 0$ for every simplex $\sigma$, hence $\partial_n \circ \partial_{n+1} = 0$ on all of $C_n$ by linearity.

This says the alternating sum of the number of cells (a combinatorial count, easy to compute by hand) equals the alternating sum of Betti numbers (a topological invariant that only depends on the shape, not the triangulation). It generalizes $V - E + F = 2$ to every dimension.

Fix $n$ and consider the boundary map $\partial_n : C_n \to C_{n-1}$ as a linear map of finite-dimensional vector spaces (working over $\mathbb{Q}$ for simplicity). By the rank-nullity theorem, $\operatorname{rank} C_n = \dim \ker \partial_n + \operatorname{rank} \partial_n$, where $\operatorname{rank}\partial_n := \dim \operatorname{im} \partial_n$.

Also, $\dim H_n(X) = \dim \ker \partial_n - \dim \operatorname{im} \partial_{n+1}$ directly from the quotient definition $H_n = \ker \partial_n / \operatorname{im}\partial_{n+1}$, so $\dim \ker \partial_n = \beta_n + \operatorname{rank}\partial_{n+1}$.

Substituting, $\operatorname{rank} C_n = \beta_n + \operatorname{rank}\partial_{n+1} + \operatorname{rank}\partial_n$. Now form the alternating sum over all $n$ from $0$ to the top dimension $N$: $\sum_n (-1)^n \operatorname{rank} C_n = \sum_n (-1)^n \beta_n + \sum_n (-1)^n \operatorname{rank}\partial_{n+1} + \sum_n (-1)^n \operatorname{rank}\partial_n$.

In the last two sums, the term $\operatorname{rank}\partial_k$ appears once from the $\operatorname{rank}\partial_n$ sum (at $n=k$, with sign $(-1)^k$) and once from the $\operatorname{rank}\partial_{n+1}$ sum (at $n=k-1$, with sign $(-1)^{k-1}$); these two signs are opposite, so every $\operatorname{rank}\partial_k$ cancels exactly (telescoping), except the boundary terms $\partial_0 = 0$ and $\partial_{N+1}=0$ which contribute nothing anyway.

What remains is $\sum_n (-1)^n \operatorname{rank} C_n = \sum_n (-1)^n \beta_n$, and since the left side is $\chi(X)$ by definition, this is exactly $\chi(X) = \sum_{n\ge 0} (-1)^n \operatorname{rank} C_n = \sum_{n\ge 0} (-1)^n \beta_n(X)$.