In algebraic topology, we define a degree for a map $f: S^n \to S^n$ as where the induced map $f\_\*$ on the $n$-th homology group of $S^n$ sends $1$.


In differential topology, we have a different (same?) notion of degree for $f$. You take a regular value $b \in S^n$, consider $f^{-1} (b)$ (which is finite by the inverse function theorem and some compactness argument), and take the difference between the number of points in the preimage where the Jacobian of $f$ is positive and the number of points in the preimage where the Jacobian of $f$ is negative.


Geometrically, I can see that they are the same, but I couldn't convince myself rigorously. In Prop 2.30 of Hatcher, he mentions that the degree of $f$ is the sum of the local degrees of $f$ at each preimage point, and local degrees are either $\pm 1$. (Local degree is defined in the middle of page 136 in Hatcher.)


So, the final question is, must the sign of the local degree of $x \in f^{-1}(b)$ the same as the sign of the Jacobian of $f$ at $x$?



Notions of degree for maps $S^n \to S^n$?
I think what you need is the following lemma (usually called the "Stack of records" lemma):


Consider a smooth proper map of manifolds of the same dimension $f \colon M \to N$ and let $y \in N$ be a regular value of $f$.


Then there exists a neighbourhood $V \subset N$ of $y$ such that $f^{-1}(V) = \cup\\_{i=1}^n U\\_i$ with $U\\_i \cap U\\_j = \emptyset$ for $i \neq j$ and $f|\\_{U\_i} \colon U\\_i \to V$ is a diffeomorphism for all $i$.


  
  



Now from this you can just sum up $\pm 1$ according to orientation on each $U\_i$ to get the local degree of $f$ at $y$, and this works for both definitions of degree.


