Given that $$\Delta_n=(x_1,...,x_{n+1})\in \{\mathbb R^{n+1}: \Sigma x_i=1, x_i\geq0\},$$ set $P=\Delta_m\times\Delta_n$ and $s=m+n$. Since each point of $P$ and $\Delta_s$ can be describe using $s$ parameters, they can be embedded in $\mathbb R^s$ as $P'$ and $\Delta_s'$. Assume, wlog, $P'\subset\Delta_s'$ (scaling and translation are homeomorphisms). Now, let $a$ be a point in the interior of $P'$ and $\partial P'$, $\partial\Delta_s'$ the boundaries of theses sets. Furthermore, for all $x\in P'\backslash\{a\}$, let $r_x$ be the ray that starts at $a$ and passes through $x$. Then, define $h:P'\rightarrow\Delta_s'$ as $h(a)=a$ and $$h(x)=a+\frac{|x'-a|}{|x''-a|}(x-a),$$ where $x'=r_x\cap \partial\Delta_s'$ and $x''=r_x\cap \partial P'$. I'd like to check that $h$ is a continuous function.

My attempt: Since $r_x=\{a+\lambda(x-a):\lambda\in \mathbb R_+\},$ $$h(x)=a+\frac{|\lambda_1(x-a)|}{|\lambda_2(x-a)|}(x-a)=a+k(x-a),$$ where $k=\lambda_1/\lambda_2$. But $k$ depends on $x$, so I guess I have to describe $\lambda_1$ and $\lambda_2$ in terms of $x$, but I could not find them. Any help will be valuable. Thank you!

As an illustration, the figure below represents $P'$ and $\Delta_s'$, if $P=\Delta_2\times\Delta_1$ (then, $\Delta_s=\Delta_3$).

enter image description here

  • $\begingroup$ I noticed we can assume $a=0$ by translation. That simplifies $h(x)$. $\endgroup$ – rgm Apr 6 '17 at 11:19
  • $\begingroup$ I could use that "Any compact convex set $K\subset \mathbb R^n$ with nonempty interior is homeomorphic to a ball". But a proof of this seems more complex to me. I found one in amakelov.github.io/2016/01/18/…. $\endgroup$ – rgm Apr 6 '17 at 14:08
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    $\begingroup$ Do you assume $P'$ and $\Delta_s'$ to be convex? You can always embed your sets in a non-convex way if you want to. // All that matters seems to be having a common interior point ($a$) for two compact convex sets in $\mathbb R^s$. $\endgroup$ – Joonas Ilmavirta Apr 6 '17 at 20:03
  • $\begingroup$ I guess that $\Delta_s'$ could be convex using the embedding $(x_1,x_2,...,x_{s+1})\mapsto (x_1,x_2,...,x_{s})$ and that $P'$ can be convex by a similar map... $\endgroup$ – rgm Apr 6 '17 at 21:39
  • $\begingroup$ Yes, that would make them convex. My point was that the sets are not convex for any embedding, but reasonable embeddings like the ones you propose do make them convex. $\endgroup$ – Joonas Ilmavirta Apr 6 '17 at 21:45

I could not answer my question exactly, but I found a way to get what I expected by it. My intent was to prove that $\Delta_n\times\Delta_m \simeq \Delta_{m+n}$. For that, I think we can use the following steps:

$(1)$ Define $f:\Delta_{m+n}\rightarrow \mathbb R^{n+m}$ as $(x_1,...,x_{m+n},x_{m+n+1})\mapsto (x_1,...,x_{m+n})$. It is easy to see that $f$ is a continuous (with $\delta:=\epsilon$, e.g.) and injective map. Since $\Delta_{m+n}$ is compact and $A:=Im(f)$ is Hausdorff, $\Delta_{m+n} \simeq A$.

$(2)$ Similarly, setting $g:\Delta_n\times\Delta_m \rightarrow \mathbb R^{n+m}$ as $$(x_1,...,x_{n+1},y_1,...,y_{m+1})\mapsto (x_1,...,x_n,y_1,...,y_m),$$ we conclude $\Delta_n\times\Delta_m \simeq B:=Im(g)$.

$(3)$ Since compactness is topological invariant, $A$ and $B$ are compact, and they are also convex and with nonempty interior. But a compact and convex set $C\subset \mathbb R^{p}$ with nonempty interior is homeomorphic to a closed $p$-ball. Then, $A \simeq B \Rightarrow \Delta_n\times\Delta_m \simeq \Delta_{m+n}$.

A proof of the affirmation in the penultimate phrase can be found in Glen Bredon's Topology and Geometry ($1993$, page $56$).

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