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If $ f,g:S^1 \rightarrow S^1$ continuous maps then \begin{equation*} \deg( f\circ g)= \deg(f)\deg(g). \end{equation*} Unfortunately, i haven't made any progress in solving it. I've tried considering the lift of f,g and $f\circ g$ but i can't see how to continue. I would appreciate any hint you have in mind.

From OP's comments: the degree of $f : S^1 \to S^1$ is defined as $\tilde{f}(1) - \tilde{f}(0)$, where $\tilde{f} : [0,1] \to \mathbb{R}$ makes the following diagram commute, where the vertical arrows are projections $t \mapsto e^{2i\pi t}$: $$\require{AMScd} \begin{CD} [0,1] @>{\tilde{f}}>> \mathbb{R} \\ @VVV @VVV \\ S^1 @>f>> S^1 \end{CD}$$ (this does not depend on the choice of $\tilde{f}$).

Najib Idrissi
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Mariospkt Pkt
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    What is your definition of the degree? This is important to know in order to help.. – Peter Franek May 25 '15 at 15:41
  • Well, this is not easy to say. If this is the first time you are dealing with algebraic topology i would recommend spending a lot of time to my question. Here, you can take a look http://www.northeastern.edu/suciu/U565/U565sp10-degree.pdf – Mariospkt Pkt May 25 '15 at 15:49
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    If your definition of degree is number coming from the induced map $H_1(S^1) \to H_1(S^1)$, then note first at the homology level, $H_\bullet(fg) = H_\bullet(f)H_\bullet(g)$, which automatically implies multiplicativity of $\deg$. Any definition of degree you have will probably be isomorphic to this definition anyway. – Balarka Sen May 25 '15 at 15:49
  • I'm not sure if we are talking for the same thing @Balarka Sen. I have posted a link in a previous comment. Let me know if you had this in mind. – Mariospkt Pkt May 25 '15 at 15:53
  • I'm sorry @Peter Franek. I mean "wouldn't " – Mariospkt Pkt May 25 '15 at 15:58
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    The definition and the properties that you already know should be part of the question, because without them this cannot be answered well. – Carsten S May 25 '15 at 16:20

2 Answers2

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If you are using the "lifting definition", just consider this commutative diagram: $$ \begin{array}{ccccc} \Bbb R & \stackrel{G}{\longrightarrow} & \Bbb R & \stackrel{F}{\longrightarrow} & \Bbb R \\ \downarrow & & \downarrow & & \downarrow \\ S^1 &\stackrel{g}{\longrightarrow} &S^1& \stackrel{f}{\longrightarrow} & S^1 \end{array}. $$ The map $G$ sends an integer $k$ to $\text{deg}(g) \times k$ (it sends $1$ to $\text{deg}(g)$, $2$ to $2\times \text{deg}(g)$ and so on) and $F$ to $\text{deg}(f)\times k$. So, the composition $F\circ G$ sends $1$ to $\text{deg}(g)\times \text{deg}(f)$.

Peter Franek
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First note:

1) that two continuous maps $f,g:S^1 \rightarrow S^1$ have the same degree iff they are homotopic.

2) If $\deg(f)=n$ and $\deg(g)=m$, then $f \simeq z^n$ and $g \simeq z^m$

3) Composition of maps behaves well with respect to composition, i.e $(f \circ g)\simeq (z^n \circ z^m)$

Now suppose $\deg(f)=n$ and $\deg(g)=m$. We use 1) 2) and 3) to obtain $\deg(f \circ g)=\deg(z^n \circ z^m)$

I think this can be an easy way to answer your question without using advanced machinery.

Remark on notation: by $z^n$ I mean $(cos(t),sin(t) \in S^1 \rightarrow (cos(nt),sin(nt)) \in S^1$ ( walking around n times in the image for each one in the domain).

D1811994
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