# Why this map has no fixed point

Let $M$ a compact manifold and $X$ a nowhere vanishing continuous vector field on $M$. Then because $M$ is compact we can define its global flow $\theta : \Bbb{R} \times M \to M$. Now $\theta|_I : I \times M \to M$ defines a homotopy between the identity on $M$ and $$\theta(1,x) : M \to M.$$

Can you give the reference for this question. Because as far as I know there is a $t$ (small) such that $\theta _t$ and $X$ has the same fixed points (fixed point of $X$ means zero's of $X$). Therefore in your case it has no fixed points. But I don't know for $t=1$. – tessellation Jun 1 '13 at 5:25
I'm confused. Let $M$ be the unit circle. Can't you just define the flow "around" the circle so that at $t=1$, the given map is the identity? – Zach L. Jun 1 '13 at 5:51
The map $\theta(1,x)$ can have fixed points, indeed it can be the identity as Zach L. pointed out.
One interpretation under which "no fixed points" is true is if it refers to the homotopy itself. Indeed, there is no $x$ such that $\theta(t,x)=x$ for all $t$. This is because differentiating with respect to $t$ we would get $X(x)=0$, contradicting the assumption that $X$ is nowhere vanishing.