The definition of the degree of a map
The degree of a continuous mapping between two compact oriented manifolds of the same dimension is a number that represents the number of times that the domain manifold wraps around the range manifold under the mapping.
$\mathbb{Q}(1):$ So how the degree of a map that the number of preimages for different point in the range does not agree?
$\mathbb{Q}(2):$ In particular, consider the function $f: \mathbb{R} \to S^1$:
$$ \Psi (n) = \left\{ \begin{array}{l l} (\sin \theta, \cos \theta) & \quad \text{for $\theta \in (0, 2\pi)$}\\ (\sin 0, \cos 0) & \quad \text{otherwise} \end{array} \right.$$
$\mathbb{Q}(3):$ Finally, how about the standard one-point compactification of $\mathbb{R}$ onto $S^1$ by steregraphic projection from north pole, with north pole identified with infinity - this is degree 1 then, right?
$\endgroup$2 Answers
$\begingroup$First, in your example the domain is not compact. More importantly, what you have is an informal definition of degree. The formal definition of degree of $f: M^n\to N^n$ is the number $d$ such that the induced map $f_*: H_n(M)\cong {\mathbb Z}\to H_n(N)\cong {\mathbb Z}$ is the multiplication by $d$.
In smooth setting, there is a more intuitive (but still formal) definition, see e.g. Guillemin and Pollack "Differential Topology".
$\endgroup$ 2 $\begingroup$Strictly speaking, your example isn't relevant because the definition required compact manifolds. But the question is reasonable anyway, and the answer is that a continuous mapping (in the compact connected case) can be slightly perturbed so that almost all points have the same, finite number of pre-images, when orientations are taken into account. (The part about orientations means that pre-image points where the map reverses orientations must be counted as negative.)
$\endgroup$
