Why if $n\geq1$ and $0\leq i\leq j\leq n$ then $F_n^j\circ F_{n-1}^i=F_n^i\circ F_{n-1}^{j-1}$?
We know that an n-simplex $\mathbb{\Delta}_n$ is a subspace of $\mathbb{R}^{n+1}$ given by $\mathbb{\Delta}_n=\{(t_0,...t_n)\in \mathbb{R}^{n+1}:\sum_{i=0}^{n}t_i=1; t_0,t_1,...,t_n\geq0\}$, and also $F_n^i:\mathbb{\Delta}_{n-1}\to \mathbb{\Delta}_n$ is such that $F_n^i(t_0,...,t_{n-1})=(t_0,...,t_{i-1},0,t_i,...,t_{n-1})$ where the zero is put in position $i+1$. I have tried to prove what I want in the following way:
$(F_n^j\circ F_{n-1}^i)(t_0,...,t_{n-2})=F_n^j(F_{n-1}^i(t_0,...,t_{n-2}))=F_n^j(t_0,...,t_{i-1},0,t_i,...,t_{n-2})=(t_0,...,t_{i-1},0,t_i,...,t_{j-1},0,t_j,...,t_{n-2})$
But on the other hand I get that
$(F_n^i\circ F_{n-1}^{j-1})(t_0,...,t_{n-2})=F_n^i(F_{n-1}^{j-1}(t_0,...,t_{n-2}))=F_n^i(t_0,...,t_{j-2},0,t_{j-1},...,t_{n-2})=(t_0,...,t_{i-1},0,t_i,...,t_{j-2},0,t_{j-1},...,t_{n-2})$
So I do not know how I can conclude that $F_n^j\circ F_{n-1}^i=F_n^i\circ F_{n-1}^{j-1}$, could someone help me please, thank you very much.