Another differential topology lemma Another lemma 



(1) Why can we assume $z=f(z)=0$ and that $U$ is convex? (the coordinate domains of the manifolds can be taken to be balls?)
(2) Why is it enough to consider the special case of a reflection?
(3) Why does it follow that "the index of $v=v_0$ at $0$ is the index of $v'=v_1$ at $0$"?
 A: For (1): given a point $z$ in a manifold you can always choose a chart so that $z$ corresponds to 0.  If $f(z) \ne 0$ then (restricting $f$ to the coordinate charts where it becomes a map of euclidean spaces) you can just replace $f$ by $\tilde f = f - f(z)$ and then $\tilde f(z) = 0$.  Making this change to $f$ does not change the index of $v'$.  And yes, you may assume $U$ is convex since you can make your coordinate charts to be open balls for example.
For (2): First show that if the result holds for two diffeomorphisms $f_1$ and $f_2$ then it holds for $f_1 \circ f_2$.  This is because
$$
index(d(f_1 \circ f_2) \circ v \circ (f_1 \circ f_2)^{-1}) = index(df_1 \circ (df_2 \circ v \circ f_2^{-1}) \circ f_1^{-1}) \\
= index(df_2 \circ v \circ f_2^{-1}) = index(v).
$$
So now suppose if $f$ is not orientation preserving.  Since we are on charts we can view it as a locally defined diffeomorphism of $\mathbb R^n$.  If it is not orientation preserving then let $R$ be the reflection $\operatorname{diag}(-1,0,\ldots,0)$.  Then $f = R \circ (R^{-1} \circ f)$ where $R^{-1}\circ f = R\circ f$ is orientation preserving (since when you take determinants of the differential you'll be multiplying by the determinant of $R$ which is -1).  So if the result is true for the reflection $R$ it will be true for $f$ since you've already shown its true for the orientation preserving map $R^{-1}\circ f$.
