I'm looking for a reference for the following:


Suppose that $f\_1,f\_2\colon S^n\rightarrow S^n$ are smooth maps. Let $i\colon S^n\rightarrow \mathbb{R}^{n+1}$ be the inclusion, and suppose that $F\colon S^n\times I \rightarrow \mathbb{R}^{n+1}$ is a smooth homotopy of $i\circ f\_1$ and $i\circ f\_2$. Moreover, suppose that 0 is a regular value for $F$. Then $$\deg(f\_2)-deg(f\_1)=d(F),$$ where $d(F)$ is the Brouwer degree of $F$ (i.e., $d(F)$ is $F^{-1}(0)$ counted with signs given by the Jacobian).


When $F^{-1}(0)=\emptyset$ this follows from the classical Hopf theorem, but I'm not sure how to prove this in general (though it should be equally classical).



Calculating degree via homotopy
I believe this follows from the fact that 'bordant' maps induce the same map on homology. So, for example: suppose $W$ is an oriented $(n+1)$-manifold with boundary and $W \to \mathbb{R}^{n+1}-\{0\}$ is a map. Then the composite $\partial W \to W \to \mathbb{R}^{n+1}-\{0\}$ is trivial on $H\_n$.


In your case you'd like to apply this in the following way:


1. Remover a tubular neighborhood of $F^{-1}(0)$ to get a manifold $W$ with boundary given (with orientations) by $\overline{S^n} \amalg S^n \amalg K$ where $K$ is the sphere bundle associated to the normal bundle of $F^{-1}(0)$.
2. Observe that $F^{-1}(0)$ was a finite collection of points, so $K$ is a union of spheres. The resulting maps have degree as you indicated (since you may shrink the original tubular neighborhood until the map $F$ is well approximated by its derivative).


To prove the fact, observe that the boundary map $H\_{n+1}(W,\partial W) \to H\_n(\partial W)$ takes the fundamental class to the fundamental class, hence is surjective, so that the next map in the sequence $H\_n(\partial W) \to H\_n(W)$ is zero.


