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This was done in Hatcher's algebraic topology example 2.32. However I do not understand it at all.

An alternative approach I know of is to show that the degree for homology matches the degree for the fundamental group which requires Hurwitz map.

For Hatcher's proof, I know that the map is locally a homeomorphism because by inverse function theorem. Then I need to determine the sign of it. How to do this mathematically instead of by words?

Does the sign have anything to do with whether the map is orientation-preserving or not?

Keith
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  • There was a related question that came up the other day here for a more general case. The idea for your case is as follows: given a $z\in S^1$ if you want to compute the local degree of $f$ at $z$, an embedding of the $1$-simplex $\sigma\colon [0,1] \to S^1$ so that $z$ is in the interior of the image represents a generator of $H^1(S, S-z)$, so you want to look at how $f$ affects the orientation of this simplex. – William Mar 07 '19 at 03:20
  • Are you using singular homology or simplicial homology?@William – Keith Mar 09 '19 at 05:36
  • Singular homology. The relative chain group $C_1(S^1, S^1 -z)$ is generated by all continuous functions $\sigma\colon \Delta^1 \to S^1$ with $z$ in their image and the cycles are given by those $\sigma$ which map the boundary into $S^1 -z$. It turns out that $H_1(S^1, S^1- z)$ is generated by any path starting and ending in $S^1 - z$ and which passes through $z$ exactly once, so for this problem you can take an embedding like I mentioned in my previous comment. – William Mar 09 '19 at 14:42

2 Answers2

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If you know that $H^1_{\mathrm{dR}}(S^1)\cong H^1(S^1;\mathbb{R})$, then using the generator $d\theta$ of de Rham cohomology, for $f(\theta)=n\theta$ (on $S^1$ parameterized with $\theta$ modulo $2\pi$) we have $df=n\, d\theta$, hence the induced map $$f^*:H^1_{\mathrm{dR}}(S^1)\to H^1_{\mathrm{dR}}(S^1)$$ is multiplication by $n$. The de Rham theorem is that there is a natural isomorphism to singular cohomology, so paired with the naturality of the universal coefficient theorem, there is the following commutative diagram:$\require{AMScd}$ \begin{CD} H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\ @VVf^*V @VVf^*V @VV(f_*)^*V\\ H^1_{\mathrm{dR}}(S^1) @>\cong>> H^1(S^1;\mathbb{R}) @>\cong>> \hom(H_1(S^1),\mathbb{R})\\ \end{CD} The UCT isomorphisms come from the fact that the $\operatorname{Ext}^1$ groups are trivial. We know that $H_1(S^1)=\mathbb{Z}$ so $\hom(H_1(S^1),\mathbb{R})=\mathbb{R}$ and $(f_*)^*$ is multiplication by $n$. This implies $f_*$ itself is multiplication by $n$, and therefore $\deg f=n$.

The Hurwitz map works, too. The horizontal maps in the following are surjections: \begin{CD} \pi_1(S^1) @>>> H_1(S^1) \\ @VVf_*V @VVf_*V \\ \pi_1(S^1) @>>> H_1(S^1) \end{CD} Since $\pi_1(S^1)=\mathbb{Z}$ is already abelian, they are of course isomorphisms. The first $f_*$ is the induced map of the covering map $f:S^1\to S^1$ in the case $n\geq 1$. From covering space theory, we can see the image is $n\mathbb{Z}$ with $1\mapsto n$, hence $\deg f=n$. For $n=0$, this is a constant map and $\deg f=0$. For negative $n$, by composition with $z\mapsto z^{-1}$ (which is degree-$(-1)$, mentioned in Hatcher as property (e) of that chapter), we get a covering map, and by property (d) we get $\deg f=(-1)(-n)=n$. There are other ways of being careful here.

The Hatcher approach, Proposition 2.30, is to let $x_1,\dots,x_m$ be the preimage of a point through $z\mapsto z^n$, with $m=\lvert n\rvert$ (and assume $n\neq 0$ since constant functions have already been handled). Concretely, the preimage of $1$ is the $m$th roots of unity, $1,e^{2\pi i/m}, e^{2\pi i\cdot 2/m}, \dots, e^{2\pi i(m-1)/m}$. Using $\theta$ coordinates, each has a an open neighborhood $U_k=(\pi (2k-1)/m, \pi (2k+1)/m)$ for $k=1,2,\dots,m$, and these are disjoint. The images of these through $z\mapsto z^n$ map such a neighborhood onto $(-\pi,\pi)$, possibly reversing orientation if $n$ is negative. The group $H_1(U_k,U_k-e^{2\pi ik/m})$ is $\mathbb{Z}$, generated by a singular $1$-simplex that crosses from one component of punctured $U_k$ to the other. The induced map $$f_*:H_1(U_k,U_k-e^{2\pi ik/m})\to H_1((-\pi,\pi),(-\pi,\pi)-0)$$ either sends this generator to a generator in the same direction, or in the opposite direction, depending on the sign of $n$ (and if you want to be careful here, consider the naturality of the long exact sequence of both pairs). Hence the local degree is $\deg f|e^{2\pi ik/m}=\operatorname{sgn} n$. Applying the proposition, $$\deg f=\sum_{k=1}^m \deg|e^{2\pi ik/m} = \sum_{k=1}^m\operatorname{sgn} n=n.$$

one potato two potato
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Kyle Miller
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  • why must $f_{\ast}$ sends a generator to a generator? – Keith Mar 09 '19 at 05:35
  • @Keith The map $z\mapsto z^n$ is a homeomorphism for these open sets. It's probably best to see this using explicit $e^{i\theta}$ coordinates, while knowing that the exponential is (locally) a homeomorphism. – Kyle Miller Mar 09 '19 at 05:44
  • Right, I mentioned that in my question, but I mostly want to see how to determine the sign, sorry for the confusion. I think Hatcher is trying to show locally it has to be homotopic to the identity(when $n$ positive), so the sign has to be positive. – Keith Mar 09 '19 at 05:56
  • @Keith Maybe a way to understand Hatcher here is that he is contracting the domain $S^1$ outside $U_k$ to a point, then he calculates the degree there (relying on the fact that this is homotopic to the identity), which should then correspond to the original situation of local degree by excision. However: take some time to calculate what $f$ does explicitly to $H_1$ of these relative homology groups, they should not be mysterious. Remember, all $f_*$ does is take a singular simplex and compose it with $f$. You could even give $U_k$ and $(-\pi,\pi)$ simplicial complexes to calculate it. – Kyle Miller Mar 09 '19 at 08:39
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It is obvious that the degree of the identity map is 1. Then use the fact that the map $x\rightarrow x^n$ is homotopic to the identity map concatenated with itself n times. Lemma 4.60 in Hatcher says that for any homology theory, in particular singular homology, the induced map distributes over this sum, so the degree of $x \rightarrow x^n$ is the integer corresponding to the map $1_\mathbb{Z} + \dots +1_\mathbb{Z}:\mathbb{Z} \rightarrow \mathbb{Z}$ which is n. So the degree is n.

Connor Malin
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  • How do you show that $x \to x^n$ is homotopic to the identity map concatenated with itself n times? – Keith Mar 08 '19 at 14:58
  • One way is using complex analysis and calculate the winding number. This is just doing an integral. You should also be able to write out an explicit homotopy if you wanted. – Connor Malin Mar 08 '19 at 16:20
  • Can the homotopy be $e^i{((1-n)t+n)\theta}$? – Keith Mar 09 '19 at 05:33
  • The homotopy will have a piecewise formula. Do you remember proving the associativity of the product in the fundamental group? It will be similar to that. Each loop is split up into n sections that are just parametrized so they take a different length of time. – Connor Malin Mar 09 '19 at 08:33