I have been struggling to get a better grasp of the overall picture of what's going on throughout my solving process, perhaps, it's due a lack of a better understanding of some of the concepts involved such as fibers of an application as equivalence classes an orbits of a group action on a set also as equivalence classes and how do they relate to each other.
I was given an action of $]0, \infty[$ onto $\mathbb{R}^{n+1} \setminus\{0\}$ defined as $t\cdot x = tx$ where $t \in \hspace{0.5mm}]0, \infty[$ and $x \in \mathbb{R}^{n+1} \setminus\{0\}$, and an application
\begin{align*} \varphi\colon\hspace{0.5 mm}\mathbb{R}^{n+1} \setminus\{0\} & \longrightarrow\mathbb{S}^n \\[0ex] x & \longmapsto \frac{x}{||x||} \end{align*}
where $\mathbb{S}^n$ is the $n$-sphere. And I must prove the following: $$(\mathbb{R}^{n+1} \setminus\{0\})/{]0, \infty[}\cong\mathbb{S}^n$$
My initial thoughts on this were to begin to prove that $\varphi$ is a group morphism, which is, because $\varphi(1)=\frac{1}{||1||}=1 \hspace{0.5mm}$ and $\varphi(xy)=\frac{xy}{||xy||}=\frac{x}{||x||}\cdot\frac{y}{||y||}=\varphi(x)\varphi(y)$. And then having in mind the
First Isomorphism Theorem,
finding the kernel of $\varphi$ and checking if $\varphi$ is an epimorphism so that $Im(\varphi)=\mathbb{S}^n$ in order to assure
$$(\mathbb{R}^{n+1} \setminus\{0\})/{\ker\varphi}\cong\mathbb{S}^n$$
and hopefully having $\ker\varphi = ]0, \infty[$ but it turns out we have a monomorphism which means I can't use this line of thought. On the other hand I'm aware that we can define with the action operation with $x,y \in \mathbb{R}^{n+1} \setminus\{0\}$ as $$x\sim_{]0,\infty[}y\hspace{2mm} \text{if}\hspace{2mm} \exists\hspace{0.5mm}t\in \hspace{0.5mm}\mathbb{R}^{n+1} \setminus\{0\}: y = t \cdot x $$
and that its equivalence classes $[x]_{\sim_{]0,\infty[}}$ are the orbits of each element of $]0, \infty[$, given by
$$Orb(x) = \{y \in\mathbb{R}^{n+1} \setminus\{0\}\hspace{1mm}\colon\hspace{1mm} y = tx\}$$ besides I believe we can define another equivalence relation on $\mathbb{R}^{n+1} \setminus\{0\}$ as
$$x\sim_{\varphi} y \hspace{2.5mm}\text{if}\hspace{2.5mm} \varphi(x)=\varphi(y)$$ whose equivalence class of a particular element is a fiber of $\varphi$ given by $$[x]_{\sim_{\varphi}} = \varphi^{-1}(\varphi(x)) = \{y \in\mathbb{R}^{n+1} \setminus\{0\}\hspace{1mm}\colon\hspace{1mm} \varphi(x)=\varphi(y) \Leftrightarrow \frac{x}{||x||} = \frac{y}{||y||}\} = \{y \in\mathbb{R}^{n+1} \setminus\{0\}\hspace{1mm}\colon\hspace{1mm}y = x \cdot\frac{||y||}{||x||}\}$$ Let $t = \frac{||y||}{||x||}$ and we have that $Orb(x) = [x]_{\sim_{\varphi}}$
Is this enough to conclude that $(\mathbb{R}^{n+1} \setminus\{0\})/{]0, \infty[}\cong\mathbb{S}^n$?
I believe I'm not getting a full grasp on this and can't explain why either... My professor is giving lectures we're not supposed to have in first place, this is an 2º year undergrad exercise and I'm not even that much into abstract algebra yet, so not everything seems intuitive at first sight.
Thank you so much in advance.