[proofplan] We verify the identity $\delta_j \circ \delta_i = \delta_i \circ \delta_{j-1}$ for $i < j$ by computing the action of both sides on an arbitrary point of $\Delta^{n-2}$. Each face map $\delta_k: \Delta^{n-2} \to \Delta^{n-1}$ inserts a zero in the $k$-th coordinate. The two compositions insert zeros at positions $i$ and $j$ in different orders, but the resulting tuple is the same because the second insertion accounts for the shift caused by the first. [/proofplan]

[step:Recall the definition of the face maps] The $k$-th face inclusion $\delta_k: \Delta^{n-1} \to \Delta^n$ is the affine map that sends the vertex $e_\ell$ of $\Delta^{n-1}$ to the vertex $e_\ell$ of $\Delta^n$ if $\ell < k$, and to $e_{\ell+1}$ if $\ell \ge k$. Equivalently, for a point $(t_0, \ldots, t_{n-1}) \in \Delta^{n-1}$, the map $\delta_k$ inserts a zero in position $k$: \begin{align*} \delta_k(t_0, \ldots, t_{n-1}) = (t_0, \ldots, t_{k-1}, 0, t_k, \ldots, t_{n-1}). \end{align*} [/step]

[step:Compute $\delta_j \circ \delta_i$ on a general point of $\Delta^{n-2}$] Let $(t_0, \ldots, t_{n-2}) \in \Delta^{n-2}$. Applying $\delta_i$ first (with $\delta_i: \Delta^{n-2} \to \Delta^{n-1}$) inserts a zero at position $i$: \begin{align*} \delta_i(t_0, \ldots, t_{n-2}) = (t_0, \ldots, t_{i-1}, 0, t_i, \ldots, t_{n-2}). \end{align*} This is a point in $\Delta^{n-1}$ with $n$ coordinates. Applying $\delta_j: \Delta^{n-1} \to \Delta^n$ inserts a zero at position $j$ in this $n$-tuple. Since $i < j$, the zero already inserted at position $i$ occupies a position before $j$, so the entries at positions $j, j+1, \ldots$ in the intermediate tuple are $t_{j-1}, t_j, \ldots, t_{n-2}$. Inserting a zero at position $j$ gives: \begin{align*} \delta_j \circ \delta_i(t_0, \ldots, t_{n-2}) = (t_0, \ldots, t_{i-1}, 0, t_i, \ldots, t_{j-2}, 0, t_{j-1}, \ldots, t_{n-2}). \end{align*} This is a tuple in $\Delta^n$ with $n+1$ coordinates, having zeros at positions $i$ and $j$. [/step]

[step:Compute $\delta_i \circ \delta_{j-1}$ on the same point and verify equality]Applying $\delta_{j-1}: \Delta^{n-2} \to \Delta^{n-1}$ first inserts a zero at position $j-1$: \begin{align*} \delta_{j-1}(t_0, \ldots, t_{n-2}) = (t_0, \ldots, t_{j-2}, 0, t_{j-1}, \ldots, t_{n-2}). \end{align*} This is a point in $\Delta^{n-1}$ with $n$ coordinates. Now applying $\delta_i: \Delta^{n-1} \to \Delta^n$ inserts a zero at position $i$. Since $i < j - 1 + 1 = j$, position $i$ is strictly before the zero that was just inserted (which sits at position $j - 1$ in the intermediate tuple, but after the insertion of a zero at position $i < j-1$, it shifts to position $j$). The result is: \begin{align*} \delta_i \circ \delta_{j-1}(t_0, \ldots, t_{n-2}) = (t_0, \ldots, t_{i-1}, 0, t_i, \ldots, t_{j-2}, 0, t_{j-1}, \ldots, t_{n-2}). \end{align*} Comparing with the computation of $\delta_j \circ \delta_i$, the two $(n+1)$-tuples are identical. Since the point $(t_0, \ldots, t_{n-2}) \in \Delta^{n-2}$ was arbitrary, we conclude $\delta_j \circ \delta_i = \delta_i \circ \delta_{j-1}$ as maps $\Delta^{n-2} \to \Delta^n$.[/step]

[guided]The intuition is straightforward: both compositions insert zeros at positions $i$ and $j$ in the final $(n+1)$-tuple, but they do so in different orders. When we compute $\delta_j \circ \delta_i$, we first insert a zero at position $i$ (getting an $n$-tuple), then insert a zero at position $j$ (getting an $(n+1)$-tuple). Since the first insertion shifted everything at position $\ge i$ to the right by one, the second insertion at position $j$ lands exactly where we want. When we compute $\delta_i \circ \delta_{j-1}$, we first insert a zero at position $j-1$ (getting an $n$-tuple), then insert a zero at position $i$ (getting an $(n+1)$-tuple). Why $j - 1$ instead of $j$? Because after we insert a zero at position $i < j-1$ in the second step, everything at position $\ge i$ shifts right by one, moving the zero from position $j-1$ to position $j$. This is precisely the index shift that makes the two compositions agree. Both produce the same $(n+1)$-tuple: the original entries of $(t_0, \ldots, t_{n-2})$ with zeros inserted at positions $i$ and $j$.[/guided]