Let $\mathfrak{g}$ be a Kac-Moody algebra, $W$ be its Weyl group, generate by fundmental reflections {$r_1,...,r_n$}. For $i,j$, we have relation $(r_ir_j)^{m_{ij}}=1$,1$\le i,j\le n$. Let $W^{'}$ be a Coxeter group generate by reflections {$r_1,...,r_n$}, with relation {$(r_ir_j)^{m_{ij}}=1$,1$\le i,j\le n$}. We want to show $W\cong W^{'}$.

In Kac's book"Infinite dimensional Lie algebras" chapter 3 exercise 3.10, there is a geometric approaching as follow :Let $\pi : W'\to W$ be the canonical homomorphism (clearly it is epi). We construct a topological space $U=W'\times C/(\sim) $, where $W'$ is equipped with the discrete topology, the fundamental chamber $C$ with the metric topology, and $\sim$ is the following equivalence relation: $(w_1,x)\sim (w_2,y) \Leftrightarrow \{x=y\ and\ w_1^{-1}w_2\ lies\ in\ the\ subgroup\ of\ W'\ generated\ by\ those\ r_i\ which\ fix\ x\}$. Define an action of $W'$ by: $w(w_1,x)=(ww_1,x)$. This is well defined. Then there exsit a continuous $W'$-equivariant map $\phi :U\to X$( where $X$ is the Tits cone  ), such that $\phi(1,x)=x$ for $x\in C$ (  $W'$operates on $X$ via $\pi$). This map is just $\phi(w,x)=\pi (w)x$. Let $Y=\{x| x\ is\ fixed\ by\ at\ least\ three\ fundamental\ reflections\ from\ W \}$ and set $X'=X\backslash Y$, $U'=U\backslash \phi^{-1}(Y)$. I can show that $\phi': U'\to X'$ is a cover map with cover multiplicity $|Ker(\pi)|$(where $\phi'$ is the restrict of $\phi$).

My question is how to deduce that $\phi'$ is a homeomorphism ? If so, then $\pi$ is a isomorphism.

Let $\mathfrak{g}$ be a Kac-Moody algebra, $W$ be its Weyl group, generate by fundmental reflections {$r_1,...,r_n$}. For $i,j$, we have relation $(r_ir_j)^{m_{ij}}=1$, $1\le i,j\le n$. Let $W^{'}$ be a Coxeter group generate by reflections {$r_1,...,r_n$}, with relation {$(r_ir_j)^{m_{ij}}=1$, $1\le i,j\le n$}. We want to show $W\cong W^{'}$.

In Kac's book"Infinite dimensional Lie algebras" chapter 3 exercise 3.10, there is a geometric approaching as follow :Let $\pi : W'\to W$ be the canonical homomorphism (clearly it is epi). We construct a topological space $U=W'\times C/(\sim) $, where $W'$ is equipped with the discrete topology, the fundamental chamber $C$ with the metric topology, and $\sim$ is the following equivalence relation:

$$(w_1,x)\sim (w_2,y) \quad \Leftrightarrow \quad x=y \text{ and } w_1^{-1}w_2 \text{ lies in the} $$ $$\text{subgroup of } W' \text{ generated by those } r_i \text{ which fix } x.$$ Define an action of $W'$ by: $w(w_1,x)=(ww_1,x)$. This is well defined. Then there exist a continuous $W'$-equivariant map $\phi :U\to X$  (where $X$ is the Tits cone), such that $\phi(1,x)=x$ for $x\in C$ ($W'$ operates on $X$ via $\pi$). This map is just $\phi(w,x)=\pi (w)x$. Let $$Y=\{x\mid x \text{ is fixed by at least three fundamental reflections from } W \}$$ and set $X'=X\backslash Y$, $U'=U\backslash \phi^{-1}(Y)$. I can show that $\phi': U'\to X'$ is a cover map with cover multiplicity $|\mathrm{Ker}(\pi)|$ (where $\phi'$ is the restriction of $\phi$).

My question is how to deduce that $\phi'$ is a homeomorphism ? If so, then $\pi$ is a isomorphism.