Q:

Manifold non-orientable iff. frame bundle is connected

Let $M$ be a connected smooth manifold and $L(M):=\bigcup_{x\in M}L_xM$ its frame bundle where $L_xM:=\{(v_1,\dots,v_n):\{v_1,\dots,v_n\}\text{ is a basis of }T_xM\}$.

$M$ is non-orientable iff. $L(M)$ is connected.

I think I have managed to prove, that if $L(M)$ is connected then $M$ is non-orientable by assuming $M$ is orientable and showing that then $L(M)$ has two disjoint components, namely the set of all positively oriented bases and the "rest". However, I struggle on proving the converse. My idea is to use non-orientability to show path-connectedness but I cannot construct such a continuous path in $L(M)$. 

A:

It's a fact that given a path $c:[0,1] \to M$ and a (positive) basis of $T_{c(0)} M$, you can "move" this basis to $c(1)$ (the formal definition would be $\exists$ vector fields $v_1,\dots,v_n$, where $n = \dim M$, along the path such that $\{v_i(0)\}$ coincides with your starting basis), by continuity the basis of $T_{c(1)} M$ is also positive.
Now try to prove that $M$ is orientable iff for every loop $\lambda:[0,1] \to M$, the orientation on $T_{\lambda(0)}M$ is preserved when you move it to $T_{\lambda(1)}M$ in the above sense. Can you see how your question follows? 

