The Mobius band is nonorientable although its boundary circle is orientable

Suppose $B$ had an orientation. Pulling its determinant rays back by the local diffeomorphism $q$ would orient the connected strip $\mathbb{R}\times [-1,1]$. Relative to the standard ray of $({\mathrm{\partial }}_t,{\mathrm{\partial }}_s)$, this continuous choice has one constant sign on the strip.

Since $q\circ g=q$, the pulled-back orientation would have to be invariant under $g$. But $Dg=\mathrm{diag}(1,-1)$ has determinant $-1$ and reverses every determinant ray, contradicting step 1.1. Hence $B$ is nonorientable by [L2].

The two boundary lines of the strip are exchanged by $g$, so their quotient is one component. The map $t<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>2\mapsto q(t,1)$ is a smooth bijection $\mathbb{R}\mathrm{/}2\mathbb{Z}\to \mathrm{\partial }B$ with smooth inverse in the quotient charts. It identifies $\mathrm{\partial }B$ with a circle, whose positive $t$-direction supplies an orientation under [L1]. This orientation is chosen on the boundary itself and is not induced from the nonorientable band.