# The Cartesian product of two simplices

Here is a result similar in flavor to the barycentric subdivision post I wrote:

If $\sigma$ and $\tau$ are two simplices of dimension at least 1, the product $\sigma \times \tau$ is generally not a simplex. But

(a) Suppose $\sigma = \langle v_0, \ldots, v_n \rangle$ and $\tau = \langle w_0, \ldots, w_m\rangle$, with vertices ordered as indicated. Show that for every pair of sequences $0 \le i_0 \le i_1 \le \cdots \le i_k \le n$ and $0 \le j_0 \le j_1 \le \cdots \le j_k \le m$, with $i_p < i_{p+1}$ or $j_p < j_{p+1}$ for every $0 \le p \le k-1$, the points $$(v_{i_0}, w_{j_0}), (v_{i_1}, w_{j_1}), \ldots , (v_{i_k}, w_{j_k})$$ span a $k$-simplex.

(b) Show that these simplices form a simplicial complex, whose topological space is $\sigma \times \tau$.

Based upon the response I received in the barycentric subdivision case, I should have a strategy for proving (b) from (a), but I was wondering if anyone visiting today knows how to deduce part (a), and if so, would be willing to give suggestions for deducing (a), or outlining a proof of it. Any constructive responses would be greatly appreciated!

-