I'm trying to understand the proof of theorem $3.62$ page 123 from "Foundations of differentiable manifolds and Lie groups" by Frank F. Warner
Let $\eta : G \times M \rightarrow M$ be a transitive action of a Lie group on the manifold $M$ on the left. Let $m_0 \in M$ and $H = \{g \in G : \eta(g,m_0) = m_0 \}$ be the isotropy group at $m_0$. Define a mapping $$ \tilde \beta : G/H \rightarrow M : gH \mapsto \eta(g,m_0).$$ Then $\tilde \beta $ is a diffeomorphism.
In the proof the author states that the kernel of the differential map of $\pi$ at $\sigma \in G $ is the tangent space to $\sigma H$ at $g$. That is
$$ \ker \pi_{*\sigma} = T_{\sigma}(\sigma H) \subset T_\sigma G.$$
where
$$ \pi_{*\sigma} : T_\sigma G \rightarrow T_{\sigma H} G/H $$
I don't understand why this is the case. I presume that it is obvious since the author doesn't give a detailed explanation.
My (updated) attempt (Using Thorgott's hint.)
From previous results I know that $H$ is closed which means $\sigma H$ is closed. From this I deduce that $\sigma H$ is a Lie subgroup of $G$ which is embedded in $G$.
I can therefore identify the tangent space $T_{\sigma}{\sigma H}$ with a subspace of $T_\sigma G$ by using the differential of the inclusion map.
Let $X_\sigma \in T_\sigma \sigma H$ and $X_{i(\sigma)} = i_{*\sigma} X_{\sigma} \in T_\sigma G.$ Let $f \in C^\infty (G/H)$. Then
$$ \pi_* X_{i(\sigma)} f = X_{\sigma} (f \circ \pi \circ i).$$
But $ f \circ \pi \circ i$ is constant on $\sigma H$ since $f \circ \pi \circ i(\sigma h) = f(\sigma h H) = f (\sigma H)$. Therefore
$$\pi_* X_{i(\sigma)} = 0$$ and $T_{\sigma} \sigma H \subset \ker \pi_{*\sigma}.$
Using ThorGott's hint again I want to show that $\ker \pi_{*\sigma}$ has the same dimension as $T_\sigma \sigma H = \dim H$ to show the equality.
Since $\dim T_{gH} G/H = \dim G - \dim H$ I can conclude from the rank nullity theorem that $\dim \ker \pi_* = \dim H$ only when $\pi_*$ is surjective.
How could I show surjectivity ?
Hint: The map $\pi$ is constant on $\sigma H$, which gives one inclusion. Equality then follows by counting dimensions.