Given two orientable manifold $M$ and $N$ with same dimension, $M$ being compact. If we have a smooth map
$$f:M \to N$$ let $p$ be a regular value, then in a neighborhood of each $x$ in $f^{-1}(p)$ the map $f$ is a local diffeomorphism. Diffeomorphisms can be either orientation preserving or orientation reversing. Let $r$ be the number of points $x$ at which $f$ is orientation preserving and $s$ be the number at which $f$ is orientation reversing. When the codomain of $f$ is connected, and one defines the degree of $f$ to be $r − s$.
Now my question is if we change the orientation of the manifold $M$ or $N$ won't the degree of the map change?
I don't get how the degree is well defined in this sense.
Degree of a map is well defined
differential-geometrydifferential-topology
Best Answer
The degree indeed depends on the orientations of the manifolds $M$ and $N$: If you change the orientation of either $M$ or $N$ (or both) the degree changes sign (or keeps the same, respectively).
So one should better say: The degree is defined if $M$ and $N$ are oriented (rather than just orientable). It is just laziness that one speaks about orientable only, because this is of course the crucial topological property which $M$ or $N$ should have - the choice of the actual orientiation is then just a “convention” (which one should specify, of course, if one really wants to calculate a degree of a map).