Determine the sign of local degree In Hatcher's P136, the local degree of a map $f:S^n\to S^n$ is the degree of the map $f_*:H_n(U,U-x)\to H_n(V,V-y) $ where $U$ and $V$ are neighbourhoods of $x$ and $y$ such that $f(U)\subset V$ and $f(U\setminus\{ x\})\subset V\setminus \{ y\}$. Both homology groups are isomorphic to $\mathbb{Z}$ so the degree is $\pm $ an integer. Is there a convention for determining the sign of the local degree? What will be a sensible way for choosing a generator?
 A: If you are using singular homology (as opposed to cellular or simplicial) a generator of $H_n(U, U-x) \cong H_n(\mathbb{R}^n, \mathbb{R}^n -0)$ can be taken as any embedded $n$-simplex $\varphi\colon \Delta^n \to U$ such that $x$ is in the interior of $\varphi(\Delta^n)$. The ordering of the vertices in the simplex induces a local orientation at $x$, so we can chose $\varphi$ to induce a positively oriented generator (wrt the standard orientation on $S^n$). You can then apply the map $f$ to $\varphi(\Delta^n)$ to get a new (possibly singular) simplex $(f\circ \varphi)(\Delta^n)$ representing an element of $H_n(V, V-y)$. Then the sign of the map on relative homology groups is determined by what $f$ does to the orientation on $\Delta^n$.
EDIT: Note that the local degree doesn't necessarily need to be $\pm 1$ for an arbitrary map and neighbourhoods that satisfy your point-set conditions. If $f\colon S^n \to S^n$ is the continuous function which takes the absolute value in the $(n+1)$-st coordinate, and we choose a neighbourhood $U$ of $x=e_1$ such that $f(U) = U$ it should follow that $f_*\colon H_n(U, U-x)\to H_n(U, U-x)$ is zero, even though $f$ satisfies the point-set assumption you made. A sufficient condition for the local degree of $f$ to be $\pm 1$ at a point would be that $f$ is a local homeomorphism at that point.
