kernel of the tangent map of $p\colon G\to G/H$ Let $G$ be a Lie group and $H$ a closed subgroup. Let $p\,\colon G\to G/H$ the canonical map (here $G/H$ is just considered as a set). There is a proposition saying that $G/H$ can be equipped with a smooth structure of a smooth manifold such that $p$ is a submersion. Actually, I can fully understand that $G/H$ will be a smooth manifold of dimension $\mathrm{dim}(G)-\mathrm{dim}(H)$ but I cannot see clearly why $p$ should be submersion. My definition of submersion is that the tangent maps are surjective. So, we need to prove that the image has dimension $\mathrm{dim}(G)-\mathrm{dim}(H)$. Let $T_{g}p\,\colon T_{g}G\to T_{gH}G/H$. By linear algebra, $\mathrm{dim}(G)=\mathrm{dim}(T_{g}G)=\mathrm{dim}(\ker T_{g}p)+\mathrm{dim}(\mathrm{Im}\,T_{g}p)$. Is it obvious that $\ker T_{g}p=T_{g}H$? I cannot prove it rigorously.
(H of course is a Lie subgroup since it is a closed subgroup of G).
 A: The map $\pi:G\rightarrow G/H$ is surjective.  By Sard's theorem, there is a regular point $g_0 H\in G/H$.  That is, the map $T_{g_0} \pi: T_{g_0}G\rightarrow T_{g_0H}G/H$ is surjective for at least one $g_0\in G$.
Now, let's promote this to all of $G$.
Proposition:  The map $\pi$ has constant rank.  In particular, since it has full rank at one point, it must have full rank everywhere.
Proof:  Given any $g\in H$, consider the maps $L_g:G\rightarrow G$ and $L'_g:G/H\rightarrow G/H$ given by $L_g(g') = gg'$ and $L'_g( g'H) = (gg')H$.  Then both $L_g$ and $L'_g$ are diffeomorphisms with inverses $(L_g)^{-1} = L_{g^{_1}}$ and $(L'_g)^{-1} = L'_{g^{-1}}$.
A simple computation reveals $$L'_g \circ \pi = \pi \circ L_g.$$
Taking the derivative of the main equation at the identity $e\in G$, we find $$T_{eH} L'_g \circ T_e \pi = T_g \pi \circ T_e L_g.$$  Call this the main equation.
Since $L'_g$ is a diffeomorphism, $T_{eH}L'_g$ is an isomorphism, so we can rewrite this as $T_e \pi = (T_{eH} L'_g)^{-1}\circ T_g\pi \circ T_e L_g$.  Taking $g = g_0$, the right hand side is a composition of surjective maps, so $T_e \pi$ is also surjective.
Returning to the main equation (with arbitrary $g$), we can now use the fact that $T_e L_g$ is invertible to rewrite this as $$T_{eH} L'_g\circ T_e\pi \circ (T_e L_g)^{-1} = T_g \pi.$$  This is again a composition of surjectiv maps, so is surjective.  Since $g$ is arbitrary, we are done.  $\square$
Now that we know that map is constant rank, it follows automatically that for each $g$, $\ker T_g \pi$ has the same dimension as $T_g H$ (which I'm interpreting as $T_e L_g (\mathfrak{h})$ with $\mathfrak{h} = T_e H$).  Thus, in order to show that $T_g H = \ker T_g \pi$, we only need to show that $T_g H\subseteq \ker T_g \pi$.
To that end, let $X\in T_g H$.  Then there is a $Y\in \mathfrak{h}$ with $X = T_e L_g Y$.  Then $g\exp(tY)$ is a path through $g$ with derivative $X$ at time $0$.  Note that $\exp(tY)\in H$ for all $t$.
Now we compute:  $\pi(g\exp(tY)) = (g\exp(tY))H = gH$, so $\pi \circ (g\exp(tY))$ is constant, and has derivative zero.  This means that $T_g H\subseteq \ker T_g \pi$, so we conclude that $T_g H = \ker T_g \pi$ as desired.
