Why is the following derivation on differential map between 2 Riemannian manifold valid? I want to ask a rather simple question in Riemannian geometry in which I get stunk in the proof. The statement is as follow:
Let $T:M\rightarrow N$ be a surjective map between differentiable manifolds and $X,Y \in C^{\infty}(TM)$, $\bar{X},\bar{Y} \in C^{\infty}(TN)$ such that $dT(X)=\bar{X}$ and $dT(Y)=\bar{Y}$. Then $dT([X,Y])=[\bar{X},\bar{Y}]$.
The proof is as follow (I only list out the parts up to the steps that I do not understand)
$dT([X,Y])(f)=[X,Y](f \circ T)=X(Y(f\circ T))-Y(X(f\circ T))=X(dT(Y)(f)\circ T)-Y(dT(X)(f)\circ T)$.
My question is why the third equality is correct? To me, $Y(f\circ T)=dT(Y)(f)$, how come it can be written as $dT(Y)(f)\circ T$? I am not sure if I misunderstand something important, but I would like someone to give a good explanation on it, or provide another proof if possible.
Thanks!
 A: I think the subtlety here is what $dT(X)$ actually means.  We want it to be a section of $TN$, so we want to say how it acts on functions, i.e. we want to define $dT(X)(f)\in C^\infty(N)$, for $f\in C^\infty(N)$.  
Now the natural thing to write down is $dT(X)(f)=X(f\circ T)$.  In fact we can always write this down, regardless of whether $T$ is surjective, but now we have a function on $M$, not on $N$.  So we need assumptions both on $T$ and on our vector field $X$ for this to make sense.  The condition on $T$ that we need is that $T$ is surjective, as you noted.  And on $X$, we want to say that $dT(X)$ actually makes sense as a vector field, so we need $X$ to be 'constant' along the fibers of $T$, i.e. that for any $x\in N$, $y\in T^{-1}(x)$ we need that $(dT)_y(X_y)\in T_xN$ is a fixed vector, which it looks like we are defining to be $\overline{X}_x$. (Perhaps we should additionally assume $T$ is a submersion to guarantee that $\overline{X}$ will be a smooth vector field, although I'm unsure if this is necessary).
Now, assuming these properties on $T$ and $X$, it follows that $X(f\circ T)$ will be a smooth function on $M$ which is constant along the fibers of $T$, and hence descends to a smooth function on $N$, which is how we define $dT(X)(f)\in C^\infty(N)$.  Let's make this explicit by writing, for $g$ a function on $M$ constant along the fibers of $T$, $g^{desc}$ to mean the function on $N$ we get from $g$.  So by definition, $dT(X)(f)=(X(f\circ T))^{desc}$.
Now, going back to your proof.  We are now assuming that $X$ and $Y$ both satisfy the nice properties listed above (i.e. 'constant' along the fibers of $T$).  Then the claim we want to make is:
$$dT([X,Y])(f)=([X,Y](f\circ T))^{desc}=(X(Y(f\circ T)))^{desc}-(Y(X(f\circ T)))^{desc}$$
Now, the fact that the proof uses in the step you're confused about is that 
$$(X(Y(f\circ T)))^{desc}=(X(Y(f\circ T))^{desc}\circ T)^{desc}$$
And this is obvious: if $g$ is a smooth function on $M$ constant along the fibers of $T$, then it's always true that $g^{desc}\circ T=g$.  Apply this to the function $Y(f\circ T)$.  
