Proving a Covariant Derivative is Torsion Free Let $(M,g)$ be a metric manifold and $\phi:M\to N$ a diffeomorphism, where $N$ is another manifold. Let $\nabla$ be the Levi Civita connection with respect to the metric $g$, and we define a connection in $(N,\phi_*(g))$ by:
$$\tilde{\nabla}_XY=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)$$
I am trying to prove that $\tilde{\nabla}$ is the  Levi Civita connection of $(N,\phi_{*}(g))$. So there is a step in which I evidently make a mistake and I am not being able do find it. Im trying to prove that it is torsion free (I don't want an alternative way to do this, I simply want to understand what am I doing wrong) so lets go. We want to see that
$$\tilde{\nabla}_XY-\tilde{\nabla}_YX=[X,Y]$$
So, since $\phi$ is a diffeo, we have that $\phi^*=(\phi^{-1})_*$ and thus,
\begin{equation}
\begin{split}
\tilde{\nabla}_XY-\tilde{\nabla}_YX &=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)-\phi_*\left(\nabla_{\phi^*(Y)}\phi^*(X)\right)\\
&=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)-\nabla_{\phi^*(Y)}\phi^*(X)\right)\\
&=\phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right)
\end{split}
\end{equation}
where in the second line I used linearity of $\phi_*$ and in the third line that $\nabla$ is the LC connection and thus Torsion Free. So now comes the problem, lets act on a function $f:N\to \mathbb{R}$ to get:
\begin{equation}
\begin{split}
\phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right)(f)&=\left[\phi^*(X),\phi^*(Y)\right]\left(\phi^*(f)\right)\\
&=\phi^*(X)\left\{\phi^*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{\phi^*(X)(\left(\phi^*(f)\right))\right\}\\
&=\phi^*(X)\left\{(\phi^{-1})_*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{(\phi^{-1})_*(X)(\left(\phi^*(f)\right))\right\}\\
&=\phi^*(X)\left\{Y(f)\right\}-\phi^*(Y)\left\{X(f)\right\}
\end{split}
\end{equation}
where in the second line I have used that $[X,Y](f)=X(Y(f))-Y(X(f))$, in the third that $\phi^*=(\phi^{-1})_*$ and in the last line I used the definition of the pushforward of $X$ and $Y$ and the pullback of $f$ via $\phi^{-1}$ and $\phi$ respectively. However, in this last step there should be a mistake I am not being able to find since $\phi^*(X)$ should act on functions $h:M\to \mathbb{R}$ while $Y(g):N \to \mathbb{R}$.
Any help will be appreciated.
 A: The mistake is not only in the last line but also in the first line of your last mathematical paragraph. Let $Z$ be a vector field on $M$ and $f \colon N \rightarrow \mathbb{R}$ be a smooth function. In your derivation, you are using the "identity"
$$ \left( \phi_{*} \left( Z \right) \right)(f) = Z(\phi^{*}(f)) $$
(where $Z = [\phi^{*}(Z),\phi^{*}(Y)]$) but this identity clearly cannot hold as the left hand side is a function on $N$ while the right hand side is a function on $M$.
More explicitly, for $p \in M$ and $q \in N$ we have
$$ (\phi_{*}(Z)(f))(q) = df|_{q} \left( d\phi|_{\phi^{-1}(q)} \left( Z|_{\phi^{-1}(q)} \right) \right), \\
(Z(\phi^{*}(f)))(p) = d(\phi^{*}(f))|_{p}(Z_p) = d(f \circ \phi)|_{p}(Z_p) = df|_{\phi(p)} \left( d\phi|_{p} \left( Z_p \right) \right) $$
so a correct identity is
$$ \left( \phi_{*}(Z) \right)(f) = \phi_{*} \left( Z(\phi^{*}(f)) \right) $$
(this is an identity between functions on $N$ and you can also write another one in terms of functions on $M$).
Similarly, the identity
$$ \phi^{*}(X) \left( \phi^{*}(f) \right) = X(f) $$
is wrong but
$$ \phi^{*}(X) \left( \phi^{*}(f) \right) = \phi^{*} \left( X(f) \right) $$
is correct.
Thus,
$$ \phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \right)(f) = 
\phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \left( \phi^{*}(f) \right) \right)= 
\\
\phi_{*} \left( \phi^{*}(X) \left( \phi^{*}(Y) \left( \phi^{*}(f) \right) \right) - \phi^{*}(Y) \left( \phi^{*}(X) \left( \phi^{*}(f) \right) \right) \right) =
\\
\phi_{*} \left( \phi^{*}(X(Yf)) - \phi^{*}(Y(Xf)) \right) = X(Yf) - Y(Xf) = [X,Y](f). $$
