Why the induced orientation need the manifold is connected

Question

Picture below is from 18th page of Do Carmo's Riemannian Geometry. In my view, the induced orientation does not connected. Assume $(U_\alpha, x_\alpha)$ is orientation of $M_1$, then $(\varphi(U_\alpha), \varphi\circ x_\alpha)$ is a orientation of $M_2$. Seemly, the connected is not used in here.

Answer 1

The connectedness assumption isn't required to define the induced orientation, but is instead required to make sense of of the last sentence: in the case where $M,N$ are not connected, a diffeomorphism $\varphi:M\to N$ may be neither orientation reversing nor orientation preserving.

As a general note, in many topics it's common to assume connectedness as a matter of convenience, even when it is not strictly necessary. Extending to the nonconnected case (by treating connected components individually) is usually trivial and/or not particularly useful.

Answer 2

Your statement is correct, but really not very relevant to the author's purpose of giving a meaningful definition of orientation-preserving vs orientation-reversing diffeomorphisms.

If $\varphi:M_1\to M_2$ is a diffeomorphism and say $M_1$ is not connected, notice that for any connected component $C$ of $M_1$ that the restriction $\varphi\vert_C : C\to \varphi(C)$ is a diffeomorphism between connected manifolds. The restriction can preserve or reverse orientation by the author's definition.

So for each connected component, $\varphi$ may preserve or reverse orientation. This is the usual way to discuss disconnected manifolds: Break them down into their connected components and talk about those.

Answer 3

If $M$ has $n$ different connected components and if $M$ is orientable, each connected component is orientable and $M$ has $2^n$ different possible orientations.

Thus, if $f : M_1 \to M_2$ is a diffeomorphism between oriented manifolds with finite connected components, $f$ could be

• orientation preserving ($1$ case)
• non orientation preserving ($2^n -1$ cases)

In case there is just one component, being non orientation preserving leads to only one case. In that last case, we say that $f$ is orientation reversing and this is not ambiguous which orientation is induced on $M_2$.