Prove that the tangent bundle to a smooth manifold has a canonical orientation I'm having some trouble understanding the tangent space to a point on the tangent bundle.
I have to prove for homework that the tangent bundle to a smooth manifold has a canonical orientation, even if the manifold is not orientable.
My idea is that, given bases $(v^1,\ldots,v^n)$ and $(w^1,\ldots,w^n)$ for $T_pM$ at some point $p$, these will induce two different bases $(\hat{v^1},\ldots,\hat{v^n},\ldots)$ and $(\hat{w^1},\ldots,\hat{w^n},\ldots)$ of $T_{(p,v)}TM$ for some $v \in T_pM$, where the second half of the basis is the same (because the second half of the coordinates for $TM$ are in some sense the same no matter what domain of what chart you're in, since the map is always $v\frac{\partial}{\partial x^i} \mapsto v$), and then I can use the fact that the second half is the same to show that the orientation of the two bases is the same, and hopefully it will be compatible with the smooth structure. Unfortunately, I really can't conceptualize at all what the induced basis would be like, cause I feel like I don't really have any intuition about what $T_{(p,v)}TM$ is like. Can anyone help me understand/tell me if my approach is a good one? Thanks
 A: I really don't think one can consider the orientation of the tangent bundle $TM$ canonical. What it's true is that $TM$ is always orientable. The easiest way is to exhibit a positive atlas. To do that, let $\varphi_i:U_i\to V_i\subset M$ be a familly of parametrizations of the manifold $M$, where the $U_i$'s are open in, say, $\mathbb R^n$ and $V_i$ are open in $M$ and cover $M$. Then the mappings
$$
\varPhi_i:U_i\times\mathbb R^n\to TV_i\subset TM_i:(x,u)\mapsto(\varphi_i(x),d_x\varphi_i(u))
$$
give a positive atlas of $TM$. The point is to check the jacobian determinants of the changes $H=\varPhi_j^{-1}\circ\varPhi_i$ are positive. To that end, denote $h=\varphi_j^{-1}\circ\varphi_i$, which is a diffeo. A little computation shows that
$H(x,u)=(h(x),d_xh(u))$, hence $$JH(x,u)=\begin{pmatrix}Jh(x)&0\\*&Jh(x)\end{pmatrix}.$$ (Notice that the first component of $H$ does not depend on $u$ and the second component $d_xh(u)$ of $H$ is linear with respect to $u$, hence its derivative with respect to $u$ is $d_xh$.) Consequently $\det JH(x,u)=(\det Jh(x))^2>0$.
