Considere the $X=\mathbb R\times (0, +\infty)$ endowed with the induced topology of $\mathbb R^2$. In $X$ define the equivalence relation:
$$(x_1, y_1)R(x_2, y_2)\iff x_1=x_2.$$
Is it true that the quotient $X/R$ is homeomorphic to $\mathbb R$?
I think so because if we define $f: X\rightarrow \mathbb R$ setting $f(x, y)=x$ then
$$(x_1, y_1)R(x_2, y_2)\implies f(x_1, y_1)=x_1=x_2=f(x_2, y_2)$$
so $f$ descends to the quotient $\widehat{f}: X/R\rightarrow \mathbb R$ by $$\widehat{f}([x, y])=x.$$
This map $f$ is continuous because if $\mathcal{U}$ is open in $\mathbb R$ then
$$f^{-1}(\mathcal{U})=\mathcal{U}\times (0, +\infty)$$ is open in $X$. This implies $\widehat{f}$ is continuous. Clearly $\widehat{f}$ is a bijection, but I can't finish the argument to show it is an homeomorphism.
Thanks.
Best Answer
You have the right idea, but to see the openness, we need to identify what the open sets of $X/R$ are: It consists of precisely all the sets $U\subseteq X/R$ such that the preimage of the projection onto the equivalence classes $X\to X/R$ is open.
The preimages are given precisely by sets $S\subseteq X$ which are closed under this equivalence relation, that is, if $s\in S$, then every $s'$ with $sRs'$ must be in S as well. In other words, it has to be the union of (even partitioned by) equivalence classes. In this case, equivalence classes are of the form $\{x\}\times (0,\infty)$. Therefore, S must be of the form $\mathcal S\times (0,\infty)$.
By the definition of the product topology, this is open in $X$ if and only if $\mathcal S\subseteq \mathbb R$ is open.
Thus, the map $\hat f\colon X/R\to \mathbb R$ taking $[(x,y)]=\{x\}\times (0,\infty)$ to $x$ maps an open set $\mathcal U\times (0,\infty)$ to the open set $\mathcal U$, and you're done.