Notions of degree for maps $S^n \to S^n$? - MathOverflow

Notions of degree for maps $S^n \to S^n$?

2009-11-11T08:30:15Z

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 $x$?

Answer by Elizabeth S. Q. Goodman for Notions of degree for maps $S^n \to S^n$?

2009-11-11T08:49:47Z

I think so: it looks like the local degree according to Hatcher's definition measures whether $f$ preserves orientation or reverses it on the neighborhood of $x$. On page 233 he begins discussion of orientation using excision classes: an orientation for an neighborhood in an $n$-manifold at a point $x$ is just a choice of generator of $H_n(\mathbb R^n, \mathbb R^n-x)$, and a small neighborhood $U$ about $x$ is homeomorphic to $\mathbb R^n$. In his degree-counting, he takes a neighborhood $U$ of $x$ which is disjoint from other preimages $f^{-1}(f(U))$ and looks at the sign of the map $H_n(U, U-{x})\rightarrow H_n(f(U), f(U)-{y})$.

The sign of the Jacobian of $f$ should also tell you whether $f$ is locally orientation-preserving or reversing at at $x$.

Answer by Gian Maria Dall'Ara for Notions of degree for maps $S^n \to S^n$?

2009-11-11T09:08:23Z

I think you can find a proof that the differentiable topological degree is the (co)homological degree in the book by Bott and Tu (Diffrential forms in algebraic topology). But there instead of homology they describe cohomology first. Then you need to translate everything to the homological setting (by de Rham isomorphism).

Answer by Sam Derbyshire for Notions of degree for maps $S^n \to S^n$?

2009-11-11T10:42:15Z

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.