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I have the following problem:

Let $f,g:S^1\rightarrow S^1$ be continuous mags with degree $\deg(f), \deg(g)$. Compute the degree of $g\circ f$

I first denoted $\deg(f)=n$ and $\deg(g)=m$. Then I know that f and g are homotopic to $z\mapsto z^{n,m}$ restricted to the unit circle. But I somehow don't see how to procede. Could someone help me? This would be very nice

Thank you a lot

user123234
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    Homotopy commutes with composition and preserves degree so $g\circ f \cong (z\mapsto z^m) \circ (z\mapsto z^n )= z\mapsto (z^n )^m = z\mapsto z^{mn}$ which is of degree $mn$ – user3257842 Nov 23 '21 at 19:21
  • sorry but we haven't shown all this things isn't there another possibility – user123234 Nov 23 '21 at 19:53
  • I think I quite got it, the only point I don't see yet is why Homotopy preserves degree. could you maybe explain this to me – user123234 Nov 23 '21 at 20:24
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    You can lift a map $f:S^{1}\rightarrow S^1$ to a map $\bar{f}:S^{1}\rightarrow \mathbb{R}$ such that $\bar{f}(x+1) - \bar{f}(x)$ is constant and corresponds to the degree of $f$. Applying this trough a homotopy we see that the degree of $f_{y}$ is $d(y) = \bar{f}{y}(1) - \bar{f}{y}(0)$ where $y\in[0,1]$ is the homotopy parameter. As a function of $y$ and difference of $\bar{f}_y$ this is continuous, but also only takes values in $\mathbb{Z}$, due to being a degree, so it is constant. – user3257842 Nov 23 '21 at 21:53
  • Tough I'm guessing you still need to show you can extend a lift throughout a homotopy. The exact statement is "the degree of a map is homotopy invariant", but I haven't been able to find a reference to a proof. – user3257842 Nov 23 '21 at 21:56
  • Informally speaking, you can stretch and move a rubber band around an infinite cylinder (homotopy) , but you can't change the number the times it's wrapped around (degree). – user3257842 Nov 23 '21 at 22:08
  • ah okey this makes sense with the rubber band. thank you – user123234 Nov 23 '21 at 22:28

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