# how to prove this homotopic problem

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?

• Do you mean $f\circ g$ is homotopic to $g\circ f$? – Lord Shark the Unknown Jan 13 '18 at 7:42

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 $H:I\times[0,\,1]\to \mathbb R$, $H(x,t)\to(1-t)p(x)+tq(x)$.)
• Why doesn't this prove that any two maps $S^1 \to S^1$ are homotopic? Also, if $f:S^1 \to S^1$ is $f(\cos \theta, \sin \theta) = (\cos 2\theta, \sin 2\theta)$, what are the functions $a,b : [0,2\pi) \to [0,2\pi)$? – Justin Young Jan 15 '18 at 15:16
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...