$\bar b: B/\Bbb R \rightarrow (0,+\infty) $ given by $\bar b(\bar x_1,\bar x_2)=x_1x_2$ is a homeomorphism

Question

Let $B:=[0,+\infty)\times (0,+\infty)$ I am trying to verify that the map:

$\bar b: B/\Bbb R \rightarrow (0,+\infty) $ given by

$\bar b(\bar x_1,\bar x_2)=x_1x_2$ is a homeomorphism

This map comes from passing to the quotient from the map $ b: B \rightarrow (0,+\infty) $ given by $b(x_1,x_2)=x_1x_2$ under the equivalence relation defined by the action of $\Bbb R$ on $B $ by $ t \cdot (x_1, x_2) = (e^tx_1, e^{-t}x_2)$, so I already know that is it continuous and well-defined

My try:

Surjectivity

I already proved this

Injectivity

$\bar b (\overline{(x,y)})=\bar b (\overline{(\tilde x,\tilde y)}) \implies b(x,y)=b(\tilde x, \tilde y) \implies xy=\tilde x \tilde y$ How do I conclude that $\overline{(x,y)}=\overline{(\tilde x,\tilde y)}$ ?

Continuity of the inverse:

I am also having trouble finding and proving that the inverse is continuous How should I do that? I also tried proving that the map is open or closed but no luck

I also find the notation of the quotient set a bit strange, why not $B/\sim$ as usually done in topology? I don't think $\sim =\Bbb R$

If it can help notice that h invariant under the action and that the action is free

Answer 1

Injectivity:

Suppose $x_1,x_2\in [0,\infty)$ and $y_1,y_2\in (0,\infty)$, with $b(x_1,y_1) = b(x_2,y_2) = r$. Set $t = \ln y_1 - \ln y_2$, in which case it's easily checked that $t\cdot (x_1,y_1) = (x_2,y_2)$. Thus, elements in $b^{-1}(r)$ belong to the same equivalence class mod $q$, i.e. $\overline{b}^{-1}(r)$ is a single element. Since $r$ was arbitrary, $\overline{b}$ is injective.

Continuity of the Inverse:

Consider an open set $U$ in $B/\mathbb{R}$. Above the quotient, $q^{-1}(U)$ will always be open by definition of the quotient topology. But

$$b(q^{-1}(U)) = (\overline{b}\circ q)(q^{-1}(U)) = \overline{b}(U),$$

thus we are done if we can show $b$ is an open map. This the same as showing that the arithmetic operation $\times$ is an open map on your domain.

The linked post advocates checking this "manually". One way to do this would be to consider the basis of $\mathbb{R}_{\geq 0}\times \mathbb{R}_{>0}$ consisting of open rectangles, and given any such rectangle $R:= (x_1,x_2)\times(y_1,y_2)$, show directly that $b(R) = (x_1y_1,x_2y_2)$, which is an open interval. Since your domain is half-closed the basis also includes sets of the form $[0,x_2)\times(y_1,y_2)$, which should be handled as a separate case.