Stiefel manifold is homeomorphic to $O(n)/O(n - k)$ I have a question on Stiefel Manifold
Why is the described map bijective? The statement '$A \sim B \iff AB^T \in O(n - k)$ so $V_k(\mathbb{R}^n) \cong O(n)/O(n - k)$' seems to appear from nowhere to me. 
Let's define $f: O(n) \to V_k(\mathbb{R}^n), A \mapsto (e_1, ..., e_k) A$. We have that $f(A) = f(B)$ for $A$ and $B$ in $\{ (e_1, ..., e_k)A = (e_1, ..., e_k) \ | \ A \in O(n) \} \cong O(n - k)$, so it descends to a map $\tilde{f}: O(n) \to V_k(\mathbb{R}^n)$. As $f$ is surjective, so is $\tilde{f}$, but why is $\tilde{f}$ injective? I think this is the case iff we have that $f(A) = f(B) \implies A \sim B \iff A, B \in O(n - k)$, but $f(A) = f(A^T)$ for any $A$ in $O(n)$, so this is not the case.
 A: Consider the right action of ${\rm O}(n)$ on $V_k(\Bbb R^n)$ given by $$V_k(\Bbb R^n)\times {\rm O}(n)\ni(\mathfrak{v},A)\mapsto \mathfrak{v}A\in V_k(\Bbb R^n).$$The stabilizer of a given $\mathfrak{v}\in V_k(\Bbb R^n)$ is isomorphic to ${\rm O}(n-k)$ since an element in said stabilizer is uniquely determined by it's action on ${\rm span}(\mathfrak{v})^\perp$. And the action is transitive, so the orbit-stabilizer theorem gives $V_k(\Bbb R^n)\cong {\rm O}(n)/{\rm O}(n-k)$.
As a bonus, note that we can also establish that the Grassmannian ${\rm Gr}(k,n)$ is a homogeneous space. We have the canonical map ${\rm span}:V_k(\Bbb R^n)\to {\rm Gr}(k,n)$ and a left action $${\rm O}(n)\times {\rm Gr}(k,n)\ni (A,W)\mapsto A[W]\in {\rm Gr}(k,n),$$for which the stabilizer of a given $W \in {\rm Gr}(k,n)$ is isomorphic to ${\rm O}(k)\times {\rm O}(n-k)$ (due to a block diagonal decomposition). It is also transitive, so the orbit-stabilizer theorem gives
${\rm Gr}(k,n)\cong {\rm O}(n)/({\rm O}(k)\times {\rm O}(n-k))$.
