how to prove this homotopic problem

Question

For maps $f,g: S^1\rightarrow S^1$, show that $f \circ g$ is always homtopic $f \circ g$

my friends asked me , i have no idea to solve it. could anyone help me?

Answer 1

I presume $f$ and $g$ are continuous. Let $S^1=\{(\cos\theta,\sin\theta): 0\le\theta < 2\pi\}$ and $$f(\cos\theta,\sin\theta)=(\cos a(\theta),\sin b(\theta))$$ $$g(\cos\theta,\sin\theta)=(\cos c(\theta),\sin d(\theta))$$ where $a,b,c,d:[0,\,2\pi)\to[0,\,2\pi)$ are continuous. Thus $$f\circ g(\cos\theta,\sin\theta)=(\cos a\circ c(\theta),\sin b\circ d(\theta))$$ $$g\circ f(\cos\theta,\sin\theta)=(\cos c\circ a(\theta),\sin d\circ b(\theta))$$ and $f\circ g$ is homotopic to $g\circ f$ if $a\circ c,b\circ d$ are homotopic to $c\circ a,d\circ b$ respectively. The latter is true by virtue of the fact that any two real-valued continous functions on an interval are homotopic. (For example, if $p,q$ are continuous functions on an interval $I$, a homotopy from $p$ to $q$ is $H:I\times[0,\,1]\to \mathbb R$, $H(x,t)\to(1-t)p(x)+tq(x)$.)

Answer 2

I assume you mean $g\circ f $, since that a map is homotopic to itself is trivial. .. See here for the fact that the winding number of the composition is the product of the winding numbers...

Then it follows from the fact that maps with the same winding number are homotopic...