$M$ is orientable if and only if $M \setminus \{p\}$ is orientable.

Let $M$ a manifold of class $C^{\infty}$. Show that $M$ is orientable if and only if $M \setminus \{p\}$ is orientable.

Comments:

($\Rightarrow$) Let $\omega: M \longrightarrow \Lambda^n(M)$ a orientation form of M then $\omega: M\setminus\{p\} \longrightarrow \Lambda^n(M\setminus\{p\})$ is a form which is never null then defines a orientation in $M\setminus\{p\}$.

I am having difficulty justifying the ($\Leftarrow$).

• You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p \in M$ a point, but still I'd prefer to hear that from you. – Luke Jul 23 '18 at 15:18
• For $\Leftarrow$, in a chart at $p$ write $\omega=gdx_1\wedge...\wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $\omega'=g'dx_1\wedge...\wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$. – user577471 Jul 23 '18 at 15:33
• @HGLandcaster Do you say to define $g (p) = 1$? – Croos Jul 26 '18 at 0:35
• @Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible. – user577471 Jul 26 '18 at 14:49