$T_{[g]} (G/H ) $ is isomorphic to $\mathfrak{g}/\mathfrak{h}.$ Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, and let $H$ be a Lie subgroup of $G$ with Lie algebra $\mathfrak{h}$. I want to prove that for $g \in G$,

$T_{[g]} (G/H ) $ is isomorphic to $\mathfrak{g}/\mathfrak{h}.$

To do so, I considered the map $a_{g^{-1}}: G/H \rightarrow G/H$, which associates to every class $[k] $,  the class $  [g^{-1}k].$
By differentiating this map at point $[g]$, we get the required isomorphism
$$da_{g^{-1}}|_{[g]} : T_{[g]} (G/H ) 
\rightarrow  \mathfrak{g}/\mathfrak{h}.$$
Is this true ?
 A: It is true that the map you define is an isomorphism. However, this isomorphism is not canonical. It is true that $T_{[e]}(G/H)$ can be canonically identified with $\mathfrak g/\mathfrak h$, but to define your map, you have chosen a representative $g$ of the coset $[g]$, and choosing a different representative gives a different isomorphism. What can be done canonically here is the identification of $TG/H$ as an associated bundle. Take $G\times(\mathfrak g/\mathfrak h)$ and endow it with a right $H$-action defined by $(g,X+\mathfrak h)\cdot h:=(gh,Ad_{h^{-1}}(X))$. Then $TG/H$ can be canonically identified with the space of $H$-orbits $(G\times(\mathfrak g/\mathfrak h))/H$. Edit: Explicitly, this is induced by the map that sends $(g,X+\mathfrak h)$ to $d_gp(d_e\ell_g(X))\in T_{[g]}G/H$, where $p:G\to G/H$ is the canonical map and $\ell_g$ denotes left multiplication by $g$. This can also be written as $d_[e]a_g(d_ep(X))$, so in a way it encodes the inverses of the isomorphism in the family that you constructed.
