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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?

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  • $\begingroup$ Do you mean $f\circ g$ is homotopic to $g\circ f$? $\endgroup$ – Lord Shark the Unknown Jan 13 '18 at 7:42
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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 $H:I\times[0,\,1]\to \mathbb R$, $H(x,t)\to(1-t)p(x)+tq(x)$.)

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  • $\begingroup$ 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 $a,b : [0,2\pi) \to [0,2\pi)$? $\endgroup$ – Justin Young Jan 15 '18 at 15:16
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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...

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