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Let $\Sigma_n$ and $\Sigma_m$ be two closed oriented surfaces of genus $n$ and $m$, with $n \leq m$. We may think about these surfaces as connected sums of tori, so there is an canoical inclusion map $j: \Sigma_n \to \Sigma_m$.

Furthermore, we know that the fundamental group of a closed oriented surface $\Sigma_g$ of genus $g$ is given by \begin{equation} \pi_1(\Sigma_g)= \langle x_1,y_1,\ldots,x_g,y_g \mid \prod_{i=1}^g[x_i,y_i]=e \rangle \end{equation}

My question is: how does the induced map $\pi_1(j)$ look like on fundamental groups? I do have trouble visualizing how the fundamental polygon of $\Sigma_n$ lies in the fundamental polygon of $\Sigma_m$. Any help would be appreciated.

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    $\begingroup$ What's the canonical inclusion? $\endgroup$
    – Dan Rust
    Jul 31, 2014 at 13:32
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    $\begingroup$ To echo @DanielRust, it is not surprising that you are having trouble visualizing something that does not exist. $\endgroup$
    – Lee Mosher
    Jul 31, 2014 at 13:38
  • $\begingroup$ In fact I'm pretty sure the only maps $\Sigma_n\to\Sigma_m$ with $n<m$ are homotopic to the constant map. This can be shown using a cohomology argument and considering the cup product. $\endgroup$
    – Dan Rust
    Jul 31, 2014 at 13:38
  • $\begingroup$ It's not that easy to visualize maps between surfaces of genus 2 or more. One way of generating examples is to look at congruence subgroups in arithmetic groups in SL(2,R) but basically it's a world very different from tori. $\endgroup$
    – Mikhail Katz
    Jul 31, 2014 at 14:21
  • $\begingroup$ Thank you all. Looks like I was a bit confused. @Daniel Rust: Could you sketch the argument that the only maps $\Sigma_n \to \Sigma_m$ are homotopic to the constant map or give a reference? $\endgroup$
    – Hierkommtdiemaus
    Jul 31, 2014 at 15:35

2 Answers 2

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There are no canonical inclusions among the $\Sigma_n$. On the other hand there are canonical projections from $\Sigma_n$ to $\Sigma_m$ when $n>m$. Here the induced map in $\pi_1$ simply sends the extra generators to the identity element.

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  • $\begingroup$ Thanks! How do these canonical projections look like on the level of spaces? $\endgroup$
    – Hierkommtdiemaus
    Jul 31, 2014 at 14:51
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    $\begingroup$ You "pinch off" the superfluous handles. $\endgroup$
    – Mikhail Katz
    Jul 31, 2014 at 14:54
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For at least "a little bit canonical" maps beween closed oriented surfaces, you need the contrary $n\geq m$. (you tried to imagine a map from a torus to $\Sigma_g$ which didn't work out)

(In the following, by component I mean the $i$-th part of the surface homeomorphic to a torus with $0,1$ or $2$ discs removed.)

A map $\Sigma_n \to \Sigma_m$ with $n \geq m$ could map some components canonically and some constant. In such a case you always get mapped the two generators of the $i$-th component canonically to the $j$-th component, and the two generators from the collapsed parts will be mapped to zero. This would correspond to a map $\mathbb Z^n \to \mathbb Z^m$ with corresponding matrix

$$(I_{m\times m} | 0 )$$

(note that this map has to look like this, in particular must have rank equal to $m$)

where you choose the basis of $\mathbb Z^n$ so that all canonical generators corresponding to a collapsed component of $\Sigma_n$ are appended in the end.

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  • $\begingroup$ What do you mean by "This would correspond to a map $\mathbb{Z}^n\to\mathbb{Z}^m$"? $\endgroup$
    – Dan Rust
    Jul 31, 2014 at 14:08

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