# Show that $S$ is non-orientable

Let $S$ be a regular surface covered by coordinate neighborhoods $V_1$ and $V_2$. Assume that $V_1\cap V_2$ has two connected components, $W_1$, $W_2$, and that the Jacobian of the change of coordinates is positive in $W_1$ and negative in $W_2$. Show that $S$ is non-orientable.

I know that, if a regular surface $S$, can be covered by two coordinate neighborhoods, whose intersection is connected, then the surface is orientable.

Furthermore, if $f:S\subset\mathbb{R}^3\to\mathbb{R}$ is a continuous function, in a connected surface $S$, then $f$ doesn't change of sign on $S$. Can give any hint! Thanks!

• What's your definition of orientable? Oct 5, 2016 at 2:47
• Regular surface S, is called orientable if it is possible to cover it with a family of coordinate neighborhoods. And a point $p\in S$ belongs to two neighborhoods of this family, then the change of coordinates has positive Jacobian at $p$. The choice of such a family is called and orientation of S Oct 5, 2016 at 2:51
• Possible duplicate of Manifold is not orientable Jun 20, 2019 at 17:10

Now if $S$ is orientable, then the sign for any loop must be positive (by using an oriented atlas). But if $p \in W_1$ and $q \in W_2$, and we consider a loop that starts at $p$, goes to $q$ within $V_1$, and returns to $p$ within $V_2$, then its sign must be negative.