I wonder if there is a continuous map $f\colon \Delta^{n+1}\to \Delta^n$ such that $f(x_0,x_1,...,x_n) = \left( \dfrac{x_1}{1 - x_0}, \dots, \dfrac{x_n}{1 - x_0}\right)$ if $x_0 \neq 1$.
This is an auxiliary result needed here.
I know short questions are not recieved well on this site, but I don't know what else to add.
Such a map cannot exist for $n\ge1$. The sequences $(1-1/n,1/n,0,...,0)_n$ and $(1-1/n,0,...,0,1/n)_n$ both converge to $(1,0,...,0)$, so, if such a continuous $f$ existed, we would have (since $\Delta^n$ is Hausdorff) $$f(1,\overbrace{0,...,0}^{n+1\text{ zeroes}})=\lim_{n\rightarrow\infty}f(1-1/n,1/n,\overbrace{0,...,0}^{n\text{ zeros}})=(1,\overbrace{0,...,0}^{n\text{ zeros}}),\\ f(1,\overbrace{0,...,0}^{n+1\text{ zeros}})=\lim_{n\rightarrow\infty}f(1-1/n,\overbrace{0,...,0}^{n\text{ zeros}},1/n)=(\overbrace{0,...,0}^{n\text{ zeros}},1).$$ This cannot happen unless $n=0$.
