Mixed Partials from Peter Petersen's book I am trying to understand how mixed partials are defined for a function $\gamma : \mathbb R^m \rightarrow M$, where $M$ is an $n$ dimensional manifold, from Peter Petersen's "Riemannian Geometry" (Page 112). Please refer to the book.

Let $\gamma\colon \mathbb{R}^m \to M$.  We wish to define the second partials so that they lie in $TM$ as opposed to $TTM$.
Lemma 6 (Uniqueness of mixed partials): There is at most one way of defining mixed partials so that (1) $\frac{\partial^2 \gamma}{\partial t^i \partial t^j} = \frac{\partial^2 \gamma}{\partial t^j \partial t^i}$ and (2) $\frac{\partial}{\partial t^k}g(\frac{\partial\gamma}{{\partial t^i}}, \frac{\partial \gamma}{\partial t^j}) = g(\frac{\partial^2\gamma}{\partial t^k \partial t^i}, \frac{\partial \gamma}{\partial t^j}) + g(\frac{\partial \gamma}{\partial t^j}, \frac{\partial^2\gamma}{\partial t^k \partial t^j})$ both hold.

My question is about lemma 6 (Page 112). I understand how he proves the Koszul type formula and makes an extension of $\gamma$ to $\overline{\gamma}$, but why is that $\frac{\partial^2 \gamma}{\partial t^i \partial t^j} = \frac{\partial^2 \overline{\gamma}}{\partial t^i \partial t^j}$? More specifically, what does the last line of the proof mean and how does he conclude the proof with this last statement (please refer to the link provided by Anthony below)
Thanks!
 A: As set up in the proof, suppose $\gamma: \Omega \to M$, $\Omega \subset \mathbb R^m$ is a $C^2$ map into the Riemannian manifold $M$, represented in a coordinate system by $(\gamma^1, \ldots, \gamma^n)$. We have the freedom to chose our extension, so define $$\overline \gamma : \Omega \times (-\epsilon, \epsilon) \to M$$ satisfying the conditions $\overline \gamma (x,0) = \gamma(x)$, $x \in \Omega,$ and $$\frac{\partial \overline \gamma}{\partial t^{n+1}}|_p = v \in T_p M,$$ where $p = \overline \gamma(x_0, 0)$ and $v$ are chosen arbitrarily. As demonstrated earlier in the proof, there is a Koszul-type formula which separately applies to both $\gamma$ and $\overline \gamma$. Since $\gamma(x) = \overline \gamma(x, 0),$ it follows that $\frac{\partial \overline \gamma}{\partial t^{k}}|_p = 
\frac{\partial \gamma}{\partial t^{k}}|_p$, $k < n+1$. The Koszul-type formula then gives the uniqueness statement provided that we show that the second partials of $\overline \gamma$ are equal to the second partials of $\gamma$, provided we only differentiate in the $\frac{\partial}{\partial t^k}$ direction, $k < n +1$. But $\gamma$ and $\overline \gamma$ are identically equal on the restricted domain $\Omega := \Omega \times { 0 }$, so any fixed but arbitrary definition of second partial derivative has to give the same answer on $\gamma$ and $\overline \gamma$. This is what Petersen means when he says the definition is independent of the extension.
