Let $L/K$ be a separable extension (not necessarily normal). Let $v \in K$ be a place of $K$ and let $W_{v}$ be the set of places of $L$ extending $v$. In a remark in Algebraic Number Theory by Neukirch, page 171, says the following:
Let $M$ be the normal closure of $L$ over $K$. Let $G=Gal(M/K)$ and $H=Gal(M/L)$. Let $\omega$ me any place extending $v$ to $M$. Let $G_{\omega}=G_{\omega}(M/K)$ be the corresponding decomposition group. Then, there is a bijection:
$$ G_{\omega} \setminus G / H \rightarrow W_{v}$$
given by $G_{\omega} \sigma H \mapsto \omega \circ \sigma |L$, where $G_{\omega} \setminus G / H$ is the set of double cosets $G_{\omega} \sigma H$. Thus, the number of extensions of $v$ to $L$ is the number of those cosets.
I understand why this map is surjective but I am not able to prove injectivity. Any help would be appreciated.
EDIT: I have found this question in this site Double cosets (Neukirch's Algebraic Number Theory) but I am not sure how to do it for general places. Moreover, why are the cosets swapped? Does it even matter?
$L = K(a)\cong K[x]/(f)$,
$v$ is a place of $K$,
Factorize in irreducibles $f = \prod_j f_j \in K_v[x]$
Let $M$ be the Galois closure of $L/K$ and $G = Gal(M/K),H=Gal(M/L),G_\omega = Gal(M_\omega/K_v)$ where $\omega$ extends $v$ to $M$, viewed as a subfield of $M_\omega$.
$G$ acts transitively on the roots of $f$, the stabilizer of $a$ is $H$, $G_\omega$ acts transitively on the roots of each $f_j$.
For each $j$ take some $g_j\in G$ such that $f_j(g_j(a))=0$. Then $$f_j = \prod_{b\in G_\omega g_j(a)} (x-b)= \prod_{b\in G_\omega g_j H(a)} (x-b)\in M_\omega[x] $$
$f_j(g(a))=0$ iff $g_\omega g(a)=g_j(a)$ for some $g_\omega\in G_\omega$ iff $g_\omega g\in g_j H$. So $$G_\omega g_j H = \{ g\in G, f_j(g(a))=0\}$$
ie. each $f_j$ corresponds to the double coset $G_\omega g_j H$.
The places of $L$ extending $v$ correspond to the $f_j$.