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Here is my view of orientability on a vector space $V$ of dimension $m>0$: let $I(V)$ be the set of linear isomorphisms from $V$ to $\mathbb{R}^m$. Given $\rho,\sigma\in{I(V)}$, we get a linear automorphism $\sigma\circ\rho^{-1}:\mathbb{R}^m\rightarrow\mathbb{R}^m$ with $\det(\sigma\circ\rho^{-1})\not=0$. We write $\sigma\sim\rho$ if $\det(\sigma\circ\rho^{-1})>0$, which defines an equivalence relation on $I(V)$. We define $\mathrm{Or}(V)=I(V)/\sim$, so that $|\mathrm{Or}(V)|=2$, and define an orientation on $V$ to be an element of $\mathrm{Or}(V)$. An oriented vector space is then a (finite dimensional) vector space equipped with an orientation.

Now, a basis of an oriented vector space $V$ determines an element of $I(V)$ which sends this basis to the standard basis of $\mathbb{R}^m$. We refer to this basis as positively oriented (with respect to our orientation on $V$) if this map lies in the equivalence class of our orientation, and negatively oriented otherwise.

For now I am considering an $m$-manifold embedded in $\mathbb{R}^p$. Then an orientation on such a manifold is as assignment of an orientation to each tangent space $T_x{M}$, such that there is an atlas of charts $\phi_\alpha:U_\alpha\rightarrow{V_\alpha}\subseteq\mathbb{R}^m$ such that for all $\alpha$ and all $x\in{U_\alpha}$, the map $d_x\phi_\alpha:T_x{M}\rightarrow\mathbb{R}^m$ lies in the orientation class of the orientation in $T_x{M}$. We refer to $M$ as an oriented manifold, and say $M$ is orientable if it admits an orientation.

Now, I'm aware that the following question has been asked before, but all the answers I've found have used concepts I haven't come across yet, such as differential forms, pullbacks etc. The problem is to show that if $M$ is an orientable manifold and $f:M\rightarrow{N}$ is a diffeomorphism between manifolds (where $N$ is also an $m$-manifold embedded in $\mathbb{R}^q$), then $N$ is also orientable. I've tried using the fact that $\{\phi_\alpha\circ{f}^{-1}\}_\alpha$ will be an atlas for $N$ if $\{\phi_\alpha\}$ is an altas for $M$, but am getting nowhere. I'm thinking the right way to go about it is a basis approach?

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If $g_i : {\bf R}^n\rightarrow M $ is orientable chart for $M$, then for any two, $$ {\rm det}\ d(g_2^{-1}\circ g_1 ) > 0 $$ since $M$ is orientable

So $h_i:=f\circ g_i : {\bf R}^n\rightarrow N $ is chart so that $$ {\rm det}\ d( h_2^{-1}\circ h_1) = {\rm det}\ d(g_2^{-1}\circ f^{-1} \circ f \circ g_1)= {\rm det}\ d(g_2^{-1}\circ g_1) >0 $$

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    $\begingroup$ Surely $g_i$ and $h_i$ don't have to be any chart - an orientation on a manifold only requires the existence of some chart with the above properties? $\endgroup$ – jl2 Mar 28 '16 at 10:21
  • $\begingroup$ You are right I fix it $\endgroup$ – HK Lee Mar 28 '16 at 10:26
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    $\begingroup$ Thank you - so this relies on the fact that a manifold is orientable if and only if there is an atlas that preserves orientation at the 'crossover' points? $\endgroup$ – jl2 Mar 28 '16 at 10:28

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