Symmetric connections, corrected version

My post yesterday on the torsion tensor and symmetry had a serious error. For some reason I thought that connections can be pulled back. I am correcting the latter part of that post (where I used that erroneous claim) here. I decided not to repeat the (as far as I know) correct earlier part.

Proposition 1 Let ${s}$ be a surface in ${M}$ , and let ${\nabla}$ be a symmetric connection on ${M}$ . Then $\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = \frac{D}{\partial y} \frac{\partial}{\partial x} s.\ \ \ \ \ (1)$

Assume first ${s}$ is an immersion. Then at some fixed ${p \in U}$

$\displaystyle \frac{D}{\partial x} \frac{\partial}{\partial y} s = (\nabla_{X} Y) \circ s,$

if ${X}$ is a vector field on some neighborhood of ${s(p)}$ which is ${s}$ -related to ${\frac{\partial}{\partial x}}$ , and ${Y}$ similarly for ${\frac{\partial}{\partial y}}$ . Similarly,

$\displaystyle \frac{D}{\partial y} \frac{\partial}{\partial x} s = (\nabla_{Y} X) \circ s.$

The difference between these two quantities is ${T(X,Y) \circ s=0}$ , because $[X,Y] \circ s = 0$ by a general theorem about $f$ -relatedness preserving the Lie bracket for $f$ a morphism. As before, we can approximate $s$ by an immersion at $p$ , and we get the general case.