Identifying the cotangent bundle of the flag variety Suppose $G$ is a Lie group (or I guess a linear algebraic group), $P \subset G$ a Lie subgroup with Lie algebras $\mathfrak{g}$ and $\mathfrak{p}$ respectively. In Chriss and Ginzburg's book "Representation theory and complex geometry" they have the following result (Lemma 1.4.9):

There is a natural $G$-equivariant isomorphism
  $$ T^*(G/P) \cong G \times_P \mathfrak{p}^\perp $$
  where $\mathfrak{p}^\perp$ is the annihilator of $\mathfrak{p}$ in $\mathfrak{g}^*$ and $P$ acts on $\mathfrak{p}^\perp$ by the coadjoint action.

They give a proof however I don't understand it at all. I will will reproduce it for convenience below (with some annotations that are my own!). I was hoping someone could help me fill in the details. I am actually interested in the algebraic version of the result but help understanding the proof in either the differential category or the algebraic one would be very useful!
Proof. Let $e = 1 \cdot P/P \in G/P$ be the base point. We have $T_e(G/P) = \mathfrak{g}/\mathfrak{p}$ and $T_e^*(G/P) = (\mathfrak{g}/\mathfrak{p})^* = \mathfrak{p}^\perp \subset \mathfrak{g}^*$. (NOTE: ok up to here I understand, but the rest I have no idea). It follows that, for any $g \in G$
$$T_{g \cdot e}^*(G/P) = g\mathfrak{p}^\perp g^{-1}. $$
Question 1: So I understand that somehow I am meant to be able to transfer the tangent space at $e$ to the point $g \cdot e$ using the $G$ action but I'm unsure how to write this down.
This shows that the vector bundles $T^*(G/P)$ and $ G \times_P \mathfrak{p}^\perp $ have the same fibers at each point of $G/P$, hence are equal as sets (NOTE: ok I am happy with this if I can understand the above). To prove that they are isomorphic as manifolds, one can refine the argument as follows.
Consider the trivial bundle $\mathfrak{g}_{G/P} = G/P \times \mathfrak{g}$ on $G/P$ with fiber $\mathfrak{g}$. The infinitesimal $\mathfrak{g}$-action on $G/P$ gives rise to a vector bundle morphism $\mathfrak{g}_{G/P} \longrightarrow T(G/P)$. 
Question 2: I understand that each element of the Lie algebra $X \in \mathfrak{g}$ gives rise to a vector field $\xi_X$, is the image of $(gP,X)$ under this map simply $\xi_X(gP)$?
It is clear that the kernel of this morphism is the sub bundle $E \subset \mathfrak{g}_{G/P}$ whose fiber at a point $x \in G/P$ is the isotropy Lie algebra $\mathfrak{p}_x \subset \mathfrak{g}$ at $x$. This gives an isomorphism $T(G/P) \cong \mathfrak{g}_{G/P}/E$.
Question 3: I don't know what is meant by the isotropy Lie algebra at the point but if I am correct in question 2 I would guess this is simply the tautological thing  - i.e. the elements of the lie algebra that act infinitesimally by zero?
Further, the description of the fibers of $E$ gives an isomorphism $E \cong G \times_P (\mathfrak{g}/\mathfrak{p})$. 
Question 4: I don't understand this at all
Hence, $T(G/P) \cong G \times_P (\mathfrak{g}/\mathfrak{p})$,
Question 5: Didn't we just say that $T(G/P)$ was a quotient by this bundle?
and the result follows by taking duals on each side. (NOTE: That I am happy with).
 A: I don't know much about the algebraic story so everything is a manifold for me.
The $G$-equivariant isomorphism $T^\ast (G/P) \cong G \times_P \mathfrak{p}^\perp$ is dual to the 
 $G$-equivariant isomorphism $T (G/P) \cong G \times_P (\mathfrak{g}/\mathfrak{p})$.
Denote $\pi : G \longrightarrow G/P$  the smooth natural projection defined by $\pi(g) = gP$.
 To prove the latter, consider the morphism $f : G \times (\mathfrak{g}/\mathfrak{p}) \longrightarrow T (G/P)$ defined by
 $$ (g, X + \mathfrak{p}) \mapsto d \pi_g \circ d(L_g)_e (X)$$
 where $L_g : G \longrightarrow G$ is the left multiplication. The map is well-defined since $d(L_g)_e (X) \in \ker d \pi_g$ if $X \in \mathfrak{p}$.
 For $h \in P$, $f(g h^{-1} , Ad_h (X) + \mathfrak{p}) = f(g, X) $ since $Ad_h = L_h \circ R_{h^{-1}}$ and $d \pi_{gh^{-1} } \circ d R_{h^{-1}} = d \pi_g$.
This shows that the map $f: G \times \mathfrak{g}/\mathfrak{p} \longrightarrow T(G/P) $ 
descends to a map $\bar{f} : G \times_P \mathfrak{g} /\mathfrak{p}  \longrightarrow T(G/P)$. 
$\bar{f}$ is a smooth bijection and it induces a linear isomorphism on the fibers which is already true for $f$. 
A standard lemma states that if $\phi : E \longrightarrow F$ is a morphism between vector bundles and it induces linear isomorphisms on fiber, 
then $\phi $ must be a diffeomorphism. (This follows from the fact that matrices being invertible is an open condition). 
So $\bar{f}$ is an isomorphism between vector bundles and taking the dual gives the desired result. 
