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Let $S_{g,2}$ be the orientable genus-$g$ surface with two boundary components, and let C be a simple-closed curve in $S_{g,2}$.

If C is homologically non-trivial (i.e., C does not bound a surface), and C intersects one of the boundary components, must C also intersect the other boundary component, i.e., can a non-trivial curve on $S_{g,2}$ intersect only one of the boundary components?

Edit: The question I am trying to answer is whether Dehn twists about the boundary curves are in the Torelli group, i.e., if these twists (twists in opposite directions in each boundary component) induce the identity map on homology.

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I don't understand the question. Are you asking whether an arc that starts and ends on one boundary can be homologically nontrivial? The answer to that is yes. But you did say "simple closed curve," and those can always be pushed into the interior and not intersect the boundary at all. –  Grumpy Parsnip Jul 18 '11 at 2:00
    
Yes, but I am interested in non-trivial curves that do intersect one of the components. I am trying to see whether Dehn twists in opposite directions about the boundary components cancel each other out, so that they induce the identity in homology; if the non-trivial curve passes thru both components, then twists in opposite directions would cancel each other out. Still, if all non-trivial curves can be pushed into avoiding both components, then the twists would have no effect on homology.Is this last always possible, i.e., can any non-trivial curve be made to avoid the boundary components? –  gary Jul 18 '11 at 2:13
    
By the twists cancelling each other out, I mean that the effect of one after the other (in opposite directions to each other)on a non-trivial curve would leave the curve unchanged. –  gary Jul 18 '11 at 2:21
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You are talking about a bounding pair map, which is one of the standard generators for the Torelli group. These do indeed induce the trivial map on homology. To see this, consider a curve $\gamma$ representing a homology class. If this doesn't hit either of the twisting curves, then obviously this gets sent to itself by the BP map. On the other hand, if the curve does intersect, it must do so algebraically trivially. Basically, once it enters the region bounded by the pair, it must leave it, so that the intersections come in canceling pairs. However, these could be distributed between the two boundary components, but I claim that's irrelevant for the proof. Note that each of the two twist curves for the BP map represent the same homology class $\alpha$. Then $\gamma\mapsto \gamma+(\gamma\cdot \alpha) \alpha$, where $\gamma\cdot\alpha$ is the algebraic intersection number between the two curves. Since this is zero, $\gamma\mapsto\gamma$.

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Thanks, Jim: excellent (and key) point about considering the algebraic and not geometric intersections. Re the two curves in the bounding pair, Do you mean that the two elements are homologous, i.e., if the bounding homologous curves are C1,C2, that C1-C2~0? And is the main idea then that given that the curve $\gamma$ does not (up to homotopy) intersect neither of C1,C2 to start with, then Dehn twists about C1,C2 will not affect $\gamma$ anyway? Would you please comment or suggest refs. on the importance of C1,C2 being a bounding pair; I can't see the role this plays in the result.Thanks. –  gary Jul 18 '11 at 15:59
    
Fundamental fact: If two curves cobound an oriented surface, then they are homologous, as long as you get the orientations right. The proof is to subdivide the surface into 2-simplices which then exhibit a specific chain whose boundary is the difference of the two curves. –  Grumpy Parsnip Jul 18 '11 at 17:17
    
If your curves $\alpha_1,\alpha_2$ are not a bounding pair, you can find a curve $\gamma$ which intersects them in a single point, say at $\alpha_1$. Then doing the Dehn twists will have the effect $\gamma\mapsto \gamma\pm \alpha_1$, which is not homologically trivial. –  Grumpy Parsnip Jul 18 '11 at 17:21
    
Excellent, thanks. –  gary Jul 18 '11 at 17:27
    
Just a quick thing: what kind of refs. are there for this type of material? I have never found any source that deals with this at non-expert level. –  gary Jul 18 '11 at 17:32
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