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Let $T$ be the 2-dimensional torus and let $S$ be $T$ minus one point. Then Birman exact sequence of mapping class groups becomes an isomorphism $$ \beta: Map(S)\to Map(T)=GL(2, {\mathbb Z}). $$ It is then essentially immediate that $\beta$ preserves Thurston's classification of elements of the mapping class group into three types: $\beta(f)$ is Anosov if and only if $f$ is pseudo-Anosov, etc.

Question. Did anybody bother to record this elementary observation in the literature?

I just need a reference, since anybody who knows anything about the mapping class group knows how to prove it (in several ways). (Please, do not write proofs, I know at least 4.)

I was nearly sure that Farb and Margalit have it, but they do not. Same for Casson and Bleiler, same for Ivanov. Of course, maybe this is one of the cases when it is easier to write a proof then to find a reference.

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2 Answers 2

This is (essentially) contained in Corollary 1.3 of

Birman, Joan S. Mapping class groups and their relationship to braid groups, Comm. Pure Appl. Math. 22 1969 213–238.

A special case of this corollary (the case $m=0$ and $g=1$ and $n=1$) is that the Birman kernel map $\pi_1(T) \rightarrow \text{Mod}(S)$ has kernel the entire group $\pi_1(T)$.

It would not surprise me if the result were also contained in

Magnus, W., Uber Automorphismen uon Fundamentalgruppen berandeter Flachen, Math. Ann. 109, 1934, pp. 617-646.

But I haven't had a chance to look (the kids are screaming and I have to go supervise them).

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Thank you, Andy. When you are done with the kids: Do you know of the statement in the literature that the isomorphism $\beta$ preserves types of elements of the mapping class group? As I said, I know at least 4 proofs (all completely straightforward) but do not know a single reference. Maybe instead of the proof or a reference one should simply quote Gromov "Well-known and easy to prove...". –  Misha May 12 '13 at 21:27
    
@Misha : Ah, so that's the statement you are after! I had assumed that you wanted a reference for the fact that the Birman exact sequence becomes an isomorphism for a once-punctured torus. I don't think I've ever seen the statement you want in print, though of course everyone (suitably interpreted) knows it. My inclination would just be to do as you suggest and say that it is well-known and easy to prove. –  Andy Putman May 12 '13 at 21:57

I think the correct answer is the book "Thurston's work on surfaces", or under its original French title "Travaux de Thurston sur les surfaces", by Fathi, Laudenbach, and Poenaru; english translation by Kim and Margalit.

The explicit statement you want is in the beginning of section 1.5.

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Lee: Thank you for looking. Alas, it is not there either: They only give the classification in the torus case. I am now convinced that the answer to my question is "nobody bothered". –  Misha May 13 '13 at 16:24
    
I guess what I mean is that what I would write in this situation is: "The proof for $T$ is given in [FLP], and the proof for $S$ follows immediately from the definitions." –  Lee Mosher May 13 '13 at 17:27

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