Product of standard simplexes is homeomorphic to a standard simplex

Question

The standard $n$-simplex is the set $$\Delta_n=(x_1,...,x_{n+1})\in \{\mathbb R^{n+1}: \Sigma x_i=1, x_i\geq0\}.$$ I'd like to prove that $\prod_{i=1}^p \Delta_{m_i}$ is homeomorphic to $\Delta_m$, with $m=\sum_{i=1}^p m_i$. I noticed that it is suffices to show that $\Delta_m \times \Delta_n \simeq \Delta_{m+n}$. For this, my hint is to map $z\in \Delta_m \times \Delta_n$, which is a point of the form $$(x_1,x_2,...,x_m,1-\Sigma x_i)\times(y_1,y_2,...,y_n,1-\Sigma y_j)$$ to $$(x_1,x_2,...,x_m,y_1,y_2,...,y_n,1-\Sigma x_i-\Sigma y_j).$$ If I did not misunderstand, that map is bijective, continuous and it has continuous inverse. Then, we are done. But I'm wondering if I did some mistake, since the only proof I found about this result is the one in http://www.cs.ubc.ca/~jiang/papers/NashReport.pdf (pages 8 and 9), which is longer and more complex. If someone could help me to check my attempt, I'd be grateful. Thank you!

Answer 1

Let's try and see if your proof works for $n = m = 1$. Write

$$ \Delta_1 = \{ (t, 1 - t) \, | \, t \in [0,1] \} $$

and use the $s$-parameter for the second copy of $\Delta_1$. Then you suggest defining a map $f \colon \Delta_1 \times \Delta_1 \rightarrow \mathbb{R}^3$ by

$$ f((t,1-t), (s, 1-s)) = (t, s, 1 - t - s).$$

However, we have

$$ f((1,0), (1,0)) = (1, 1, 1 - 1 - 1) = (1,1,-1) $$

so the image of the map $f$ is not contained in the simplex $\Delta_2$ so your map can't be a homeomorphism between $\Delta_1 \times \Delta_1$ and $\Delta_2$.