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I am reading a paper on Riemann surfaces and faced a problem about one of the refernces the author gave in the exlaination of one of the results.

Here is a summary of what I am reading:

Let $X_1$ and $X_2$ be two surfaces. Let $\pi_1:\widetilde{X_1}\longrightarrow X_1$, $\pi_2:\widetilde{X_2}\longrightarrow X_2$ be their universal covering surfaces. Let $G_1$ and $G_2$ be the groups of the covering transformations respectively. Let Hom($G_1, G_2$) be the set of all homomorphisms between $G_1$ and $G_2$ and define an equivalence relation by setting $j_1\sim j_2$ if there exists an inner automorphism $A: G_2\longrightarrow G_2$ such that $j_2=A \circ j_1$. Denote the equivalence class of $j$ by $[j]$

Let $f: X_1\longrightarrow X_2$ be a continuous map. We also know by a lemma that every lifting, $\tilde{f}$, of $f$ induces a homomorphism $\tilde{f_*}: G_1\longrightarrow G_2$ defined by $$ \tilde{f_*}(g)\circ \tilde{f}=\tilde{f}\circ g .$$ We also have the following lemma


Continuous mappings $f_i:X_1\longrightarrow X_2\ , \ i=1,2$, are homotopic if and only if $[\tilde{f_1}_\ast]=[\tilde{f_2}_\ast]$.

Now let $\mathbf{C}(X_1, X_2)$ denote the set of continuous mappings $f: X_1\longrightarrow X_2$, and define an equivalence relation by homotopy.

By Lemma above, we can define an injective map \begin{eqnarray} \Phi: \mathbf{C}(X_1, X_2)/\simeq &\longrightarrow& \mathbf{Hom}(G_1, G_2)/\sim \\ &[f] \longrightarrow & [\tilde{f_*}] \end{eqnarray} Clearly this map is injective.

The question is:

The author refers to a theorem by Hopf (he does not mention the theorem) to conclude that if $X_1$ and $X_2$ are compact, then $\Phi$ is bijective.

Hopf's paper was written in German and was published 80 year ago! I do not know any German. Here is a link to Hopf's paper If any body knows what theorem it is, then their help is really appreciated. Thanks in advance.

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It's rather hard to make sense of what you are saying. Which paper are you reading? –  Igor Rivin Sep 25 '11 at 9:09
The result of Hopf is Satz I of the paper you have linked to. It says exactly that $\Phi$ is a bijection if both surfaces have genus $\geq 1$. (The condition on the genus is clearly required.) –  ulrich Sep 25 '11 at 9:57
You should ask this at, this is a homework-level question for a first semester algebraic topology class (although I realize that the paper your are reading cited an old source). Also see –  Ian Agol Sep 25 '11 at 16:22
@ulrich Thank you! –  yaa09d Sep 26 '11 at 4:08

1 Answer 1

up vote 4 down vote accepted

Since the surfaces of genus $\ge 1$ are aspherical, the homotopy types of maps are classified by their induced homomorphisms of fundamental groups. This is essentially what the final result says. (But, since the fundamental group depends on a marked point in a nontrivial way, and homotopies do not respect marked points, the precise formulation involves these equivalences.)

The injectivity of $\Phi$ is covered by the "if" part of Lemma. To prove surjectivity, identify $G_1$ and $G_2$ with fundamental groups $\pi_1(X_1,x_1)$ and $\pi_1(X_2,x_2)$. Now it suffices to prove that, for any homomorphism $h:\pi_1(X_1,x_1)\to\pi_1(X_2,x_2)$ there is a map from $X_1$ to $X_2$ which maps $x_1$ to $x_2$ and induces $h$ on the fundamental group level.

This can be proved as follows. Construct a cell division of $X_1$ consisting of the marked point $x_1$ (the only 0-dimensional cell), loops $\gamma_1,\dots,\gamma_{2g}$ (where $g$ is the genus) as 1-dimensional cells (they generate the fundamental group), and a $4g$-gon $D$ attached to these loops as the only 2-dimensional cell. Map $x_1$ to $x_2$, then map each $\gamma_i$ to a loop representing $h([\gamma_i])\in\pi_1(X_2,x_2)$, then extend this map to $D$. There is no obstruction in the last step because the boundary of $D$ represents the identity of $\pi_1(X_1,x_1)$, hence the boundary of its desired image in $X_2$ (formed by the images of the $\gamma_i$'s) represents the identity of $\pi_1(X_2,x_2)$ and hence can be extended to a map of the 2-disc.

I don't read German either, but I'm fairly sure this is what is written in the proof of Satz in the cited Hopf's paper.

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@Ivanov Thank you! –  yaa09d Sep 26 '11 at 4:07
@Ivanov Why do the surfaces have to be compact? –  yaa09d Sep 26 '11 at 4:39
They don't have to compact, you can modify the argument to work with any cell decomposition (and in any dimension). All you need is that they are Eilenberg-Maclane spaces $K(\pi,1)$. As Agol mentioned, this is textbook material nowadays. –  Sergei Ivanov Sep 26 '11 at 17:22
You are right! Actually I found it in Hatcher's book, Algebraic Topology. Thank you. –  yaa09d Sep 27 '11 at 19:10

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