The index of a nondegenerate fixed point is the sign of det(I-Df)

Choose a chart centered at $x$ and a closed ball on which the chart displacement $g$ is defined. By [F2], $g(u)=Au+R(u)$ with $A$ invertible and $R(u)=o(\mathrm{\mid }u\mathrm{\mid })$. Let $c=\mathrm{\parallel }{A}^{-1}{\mathrm{\parallel }}^{-1}>0$ and shrink the ball until $\mathrm{\mid }R(u)\mathrm{\mid }\le c\mathrm{\mid }u\mathrm{\mid }\mathrm{/}2$ there. Then $\mathrm{\mid }Au\mathrm{\mid }\ge c\mathrm{\mid }u\mathrm{\mid }$, so $g$ has no zero in the punctured ball and a positive radius $\epsilon$ in it is admissible.

For $v\in S^{n-1}$ and $t\in [0,1]$, the vector $A\epsilon v+tR(\epsilon v)$ has norm at least $c\epsilon \mathrm{/}2>0$. Its normalization is a smooth homotopy from $L_A(v)=Av\mathrm{/}\mathrm{\mid }Av\mathrm{\mid }$ to the sphere map defining the fixed-point index. Therefore [F1] and [F3] identify ${\mathrm{ind}}_x(f)$ with $\mathrm{deg}{L}_A$, using reduced degree for $n=1$.

The inverse of $L_A$ is $L_{A^{-1}}$. For $n\ge 2$, radial normalization subtracts only an outward-normal component from $Aw$ on tangent vectors $w$, then rescales by a positive scalar. Thus, in outward-normal-first sphere orientations, the orientation sign of $d{L}_A$ is $\mathrm{sign}\det A$: the ambient ordered frame $(v,w_1,\dots ,w_{n-1})$ is sent to $(Av,A{w}_1,\dots ,A{w}_{n-1})$, and deleting normal components and positive rescaling leave its determinant sign unchanged. By [F3], $\mathrm{deg}{L}_A=\mathrm{sign}\det A$. For $n=1$, $L_A(v)=\mathrm{sign}(A)v$, of reduced degree $\mathrm{sign}(A)$ directly from [F1]. Finally $A$ is conjugate to $I-D{f}_x$, so their determinant signs agree. Hence ${\mathrm{ind}}_x(f)=\mathrm{sign}\det (I-D{f}_x)\in \{+1,-1\}$. This local-coordinate proof uses no orientation of $M$ and no choice principle.