I am trying to prove this fact: $SO(n+1)/SO(n) \cong S^{n-1}$. I was told to use the map that sends an element of $g \in SO(n+1)$ to $gN$ where $N$ is the north pole.
However I think this method might be wrong since the given map is not constant on the fibres of the quotient map if we consider the action of $SO(n)$ on $SO(n+1)$ as the left action, maybe instead I should use the map which sends $g$ to $Ng$. Am I right?
The action $\rho: SO(n) \times S^{n-1} \to S^{n-1}$ defined by $\rho(A,x) = A \cdot x = A(x)$, where $A \in SO(n)$ is identified with the matrix of the orthogonal linear transformation (also denoted by $A$). This action is transitive (you should prove this).
Now, consider the subset of $SO(n)$ consisting of all elements of the form $$\begin{pmatrix}& & & & 0\\& & & & 0 \\ & \tilde A & & &\vdots\\&&&&0\\0 & 0 & \cdots & 0 & 1\end{pmatrix}$$ where $\tilde A$ is a $(n-1)\times (n-1)$ real matrix satisfying $\tilde A \tilde A ^T = \tilde A ^T \tilde A= id$ and $\det \tilde A = 1$. This set is clearly a subset of $SO(n)$ isomorphic to $SO(n-1)$ (and we identify it with SO(n-1)).
Claim: The isotropy group of $e_n \in S^{n-1}$ (north pole) is $SO(n-1)$.
Observe that $A \in SO(n-1)$ implies $A \cdot e_n = e_n$. Next suppose $A$ is some element of $SO(n-1)$ that satisfies $A \cdot e_n = e_n$. Then
$$e_n = A \cdot e_n = A (e_n) = \sum_{j=1}^n A_{jn} e_j = A_{1n}e_1 + \ldots + A_{nn} e_n $$ Linear independence of $\{e_1, \ldots, e_n\}$ implies that $A_{1n} = \cdots = A_{n-1 n } = 0$ and $A_{nn} = 1$. Moreover $AA^T = id$ implies $$(A_{n1})^2 + \ldots + (A_{nn})^2 = 1$$ and $\det A = 1$ so we also have $A_{1n} = \cdots = A_{n-1 n } = 0$. Thus, $A \in SO(n-1)$ as required. As $SO(n)$ is closed in $O(n)$ you have the desired isomorphism from the following result:
Result: Let $G$ be a topological group, $X$ a topological space and $(x,g) \mapsto g \cdot x$ a transitive left action of $G$ on $X$. Fix $x_0 \in X$, let $H = \{g \in G ; g \cdot x_0 = x_0 \}$ be its isotropy subgroup and define $\mathcal Q' : G \to X$ by $\mathcal Q'(g) = g\cdot x_0$. Then $H$ is a closed subgroup of $G$ and, if $\mathcal Q : G \to G/H$ is the canonical projection, then there exists a unique continuous bijection $\varphi: G/H \to X$ for which $\varphi \circ \mathcal Q = \mathcal Q'. $
