Theorem: For any simplicial complex $K$, the boundary operators satisfy $\partial_{n-1} \circ \partial_n = 0$ for all $n$.

It suffices to verify this on generators, i.e., on individual $n$-simplices $\sigma = [v_0, \ldots, v_n]$.

Step 1: Compute $\partial_n \sigma$

By definition: $\partial_n \sigma = \sum_{i=0}^n (-1)^i [v_0, \ldots, \hat{v}_i, \ldots, v_n]$

Each term is an $(n-1)$-simplex obtained by omitting vertex $v_i$.

Step 2: Apply $\partial_{n-1}$ to each term

For the $i$-th term $\tau_i = [v_0, \ldots, \hat{v}_i, \ldots, v_n]$: $\partial_{n-1} \tau_i = \sum_{j=0}^{i-1} (-1)^j [v_0, \ldots, \hat{v}_j, \ldots, \hat{v}_i, \ldots, v_n] + \sum_{j=i+1}^n (-1)^{j-1} [v_0, \ldots, \hat{v}_i, \ldots, \hat{v}_j, \ldots, v_n]$

The first sum runs over $j < i$ (omitting $v_j$ then $v_i$), the second over $j > i$ (omitting $v_i$ then $v_j$).

Step 3: Compute $\partial_{n-1} \circ \partial_n$

$(\partial_{n-1} \circ \partial_n) \sigma = \sum_{i=0}^n (-1)^i \partial_{n-1} \tau_i$

Expanding using Step 2: $= \sum_{i=0}^n (-1)^i \left[ \sum_{j=0}^{i-1} (-1)^j [v_0, \ldots, \hat{v}_j, \ldots, \hat{v}_i, \ldots, v_n] + \sum_{j=i+1}^n (-1)^{j-1} [v_0, \ldots, \hat{v}_i, \ldots, \hat{v}_j, \ldots, v_n] \right]$

Step 4: Reindex and collect terms

Each $(n-2)$-simplex $[v_0, \ldots, \hat{v}_j, \ldots, \hat{v}_k, \ldots, v_n]$ (with $j < k$) appears exactly twice:

• Once from omitting $v_j$ first (index $i = k$, $j < i$), with sign $(-1)^k \cdot (-1)^j = (-1)^{j+k}$
• Once from omitting $v_k$ first (index $i = j$, $k > i$), with sign $(-1)^j \cdot (-1)^{k-1} = (-1)^{j+k-1}$

The signs are opposite: $(-1)^{j+k}$ and $(-1)^{j+k-1} = -(-1)^{j+k}$.

Step 5: Cancellation

Every $(n-2)$-simplex appears twice with opposite signs, so all terms cancel: $(\partial_{n-1} \circ \partial_n) \sigma = 0$

Since this holds for all generators $\sigma$, and $\partial$ is linear, we have $\partial^2 = 0$ on all chains. ∎