If $f,g:S^{1}\rightarrow S^{1}$ maps, show that $N(f\circ g)=N(f)N(g)$, where $N(f)$ is the winding number of $f$.

We defined the winding number of $f$ to be $N(f)=\frac{1}{2\pi}(\tilde{f}(1)- \tilde{f}(0) )$ where $\tilde{f}$ is a lift of $f$. I tried to look for a lift of $f \circ g$ but couldn't manage to find anything that makes sense for this problem. Can anybody help me, please? Thank you in advance!

  • $\begingroup$ What exactly do you mean by a lift of $f$? I would have thought $\tilde{f}$ is a map $S^1 \to \mathbb{R}$ such that $f = \pi\circ\tilde{f}$ where $\pi : \mathbb{R} \to S^1$ is covering projection, but in that case $\tilde{f}$ is not defined at $0$. $\endgroup$ Mar 16, 2016 at 23:57
  • $\begingroup$ They mean to think of $f, g$ as loops in $S^1$ (i.e. precompose your $\widetilde{f}$ with $\pi$). $\endgroup$
    – Pedro
    Mar 17, 2016 at 1:01

2 Answers 2


This may be a bit "cheating", but anyway... If $n = N(f)$ and $m = N(g)$, we know from the calculation of $\pi_1(S^1)$ that $f$ is homotopic to $z \mapsto z^n$ and $g$ is homotopic to $z \mapsto z^m$ (two maps with the same winding number are homotopic). But then $f \circ g$ is homotopic to $z \mapsto (z^m)^n = z^{mn}$, which has winding number $mn = N(f)N(g)$.


Here's a sketch of the idea.

Let $p : \mathbb{R} \to S^1$ be the standard covering map with period $2\pi$. Let $f' = f \circ p : \mathbb{R} \to S^1$, and let $\widetilde{g}$ be a lift of $g$. Then because $\mathbb{R}$ is convex, $\widetilde{g}$ is homotopic rel endpoints to a constant-velocity path $\gamma : I \to \mathbb{R}$. Now let $l : I \to \mathbb{R}$ be a lift of $f' \circ \gamma$. It's easy to show that $l$ is homotopic rel endpoints to a lift of $f \circ g$. The important thing is that we have the formula that $\gamma(t) = \gamma(0) + 2\pi N(g) t$.

For each $j \in \{1,\dotsc,N(g) - 1\}$, observe that for $rj \le t \le (j+1)/N(g)$, we have $l((j+t)/N(g)) = \widetilde{f}(t) + 2\pi j$. You can use this to argue by induction that for each $j \in \{1,\dotsc,N(g)\}$, we must have $l(j/N(g)) = l(0) + 2\pi jN(f)$. In particular, setting $j = N(g)$ will give you $l(1) = l(0) + 2\pi N(g)N(f)$, and you're done.

Alternatively, if you have the machinery of $\pi_1$ at your disposal and know that $\pi_1(S^1) \cong \mathbb{Z}$, you can use the fact that a homomorphism $\phi : \mathbb{Z} \to \mathbb{Z}$ can be computed via multiplication by $\phi(1)$, and that $N(f) = f^*(1)$. Then, you have $N(f \circ g) = (f \circ g)^*(1) = f^*(g^*(1)) = N(f)N(g)$.

  • $\begingroup$ Thank you very much! Great ideas! $\endgroup$
    – Lullaby
    Mar 18, 2016 at 22:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .