I am working on the following Lemma of Milnor:
Suppose that the vector field $v$ on $U$ corresponds to $$ v' =df \circ v \circ f^{-1}$$ on $U'$ under a diffeomorphism $f: U \to U'$. The the index of $v$ at an isolated zero $z$ is equal to the index of $v'$ at $f(z)$.
Proof: Let $z=f(z)=0$ and that $U$ is convex. If $f$ preserves orientation, then, proceeding exatly as above we construct a one-parameter family of embeddings $$f_t: U \to \Bbb R^m$$ with $f_0=$identity, $f_1=f$ and $f_t(0)=0$ for all $t$. Let $v_t$ denote the vector field $df_t \circ v \circ f_t^{-1}$ on $f_t(U) \subset \Bbb R^m$, which corresponds to $v$ on $U$. These vector fields are all defined and nonzero on a sufficiently small sphere centered at $0$. Hence the index of $v=v_0$ at 0 must be equal to the index of $v'=v_1$ at $0$.
I understand the proof except the last part. Why the index of $v=v_0$ at 0 must be equal to the index of $v'=v_1$ at $0$? From where this followed? Many thanks for your help.
The self-maps of $S^{m-1}$ corresponding to $t=0$ and to $t=1$ are then homotopic, hence have the same degree, i. e. the indexes of the two vector fields are equal.