Show that there exists a smooth map $F: M \to M$ that is homotopic to the identity and has no fixed points. Question (Problem 9-5 of Lee's Introduction to Smooth Manifolds): Suppose $M$ is a smooth, compact Manifold that admits a nowhere vanishing smooth vector field. Show that there exists a smooth map $F: M \to M$ that is homotopic to the identity and has no fixed points.
My try: Let $V$ be the non-vanishing smooth vector field. Since $M$ is compact, $V$ is complete. Hence there is a global flow $\theta: \mathbb{R} \times M \to M$ such that for all $t,s \in \mathbb{R}$ and $p \in M ,\theta (t,\theta(s,p))=\theta(t+s,p)$ and $\theta(0,p)=p.$
Since $p\in M$ is a regular point of $V$, there exists a neighbourhood $U_p$ in which $V$ has coordinate representation $\frac{\partial}{\partial s^1}$ . Choose a smaller neighbourhood $V_p$ such that $\theta(t,x)=x+(t,0,\ldots)$ for all $0 \le t \le t_p.$ Since $\{V_p\}_{p \in M}$ is a open cover for $M$ and $M$ is compact, there exists a smooth subcover say $\{V_{p_1},\ldots,V_{p_n}\}$. Let $T=\min \{t_{p_1},\ldots,t_{p_n}\}.$ Then $\theta_T: M \to M$ is a smooth map which has no fixed points and $H:M \times I \to M$ given by $H(x,t)=\theta(tT,x)$ is the required homotopy to identity.
I have a little issue here. I am uncertain as how can I choose the neighborhoods $V_p$ the way I did. Intuitively it seems to be true though.
Thanks for the help!
 A: Your argument is good. Here's how I'd pick appropriate $V_p$ and $t_p$:
Fix coordinates $(x,y^1,y^2,\ldots)$ on $U_p$ such that $\partial/\partial x = V$. Since $U_p$ is a neighbourhood of $p$, it contains some coordinate ball $B_r(p)$.  Since flow solutions are unique, $\theta(t,x,y) = (x+t,y)$ is true whenever $(x,y)$ and $(x+t,y)$ are both in the coordinate chart domain; and by the triangle inequality we know $(x+t,y) \in B_r(p)$ whenever $|t|<r/2$ and $(x,y)\in B_{r/2}(p).$ Thus we can take $V_p = B_{r/2}(p)$ and $t_p = r/2$.
A: I think your intuition is right, but I think there are easier ways: W.l.o.g. we may assume that $F: M\to M$, $F(m) = \theta(1,m)$ satisfies $F(m)\neq m $ (why?). 
Then $F$ is bijective (since it has an inverse given by $G(m) = \theta(-1,m)$). Is $F$ smooth? Of course, it is just the restriction of $\theta$ to $\{1\}\times M$. Moreover $\theta|_{I\times M}$ will serve as a homotopy between the identity and $F$.
Sorry, this is maybe more suitable as a comment since I don't answer your question, but I don't have enough reputations to comment.
