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enter image description here

I know that it should be either a sphere, torus, Klein bottle, real projective plane, or a connect sum of any combination of these, but I don't know the steps in identifying what kind of surface this is.

I know there are $2$ boundary components, but I don't know how many vertices or edges there are, or how many Seifert discs. There's an even number of half-twists, so this surface is orientable.

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    $\begingroup$ @ThomasAndrews: This looks to me like something someone would see if they were studying handlebody diagrams in the context of surfaces. I suspect that what's intended is to take this picture and then glue on discs to the two boundary circles. If that's the case, then I would argue that the easiest way is to do a single handleslide to 'untangle' and then recognize it as the connected sum of two known manifolds. $\endgroup$
    – user98602
    Commented Oct 8, 2015 at 1:26
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    $\begingroup$ You can compute the Euler characteristic of the surface's core graph (I think it is 2), which will be the same as the Euler characteristic of the surface (EC is a homotopy invariant), then investigate how removing discs from closed orientable surfaces changes Euler characteristic. As Mike Miller suggests, if you attach a disk along each boundary component, you'll get a closed surface and you're done by the classification of surfaces. $\endgroup$
    – Charlie
    Commented Oct 8, 2015 at 19:57
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    $\begingroup$ Nope, the twists don't matter for the core graph (I think some refer to it as the spine of the surface). Also, 2 isn't the Euler characteristic, I made a mistake. $\endgroup$
    – Charlie
    Commented Oct 8, 2015 at 20:43
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    $\begingroup$ @Charlie Here is the picture I drew. I'm not sure what constitutes as a "band" though: i.imgur.com/8IkrknK.png So I get $6$ vertices, $9$ edges, and 0 faces, so I get an Euler characteristic of $-3$ $\endgroup$ Commented Oct 8, 2015 at 20:48
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    $\begingroup$ @morphic your spine and EC are right. As Mike Miller said, you have a little more work to do since the surface is non-orientable. $\endgroup$
    – Charlie
    Commented Oct 8, 2015 at 20:58

1 Answer 1

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With a couple of moves, if you know how to recognize it you can see that it is a twice-punctured $T^2\mathop{\#}\mathbb{R}P^2$. If you do not know how to recognize it, I have draw some pictures that might convince you.

Altering the original diagram to see its decomposition

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  • $\begingroup$ +1, but minor quibble. $S^2\setminus\{x_1,x_2\}$ is an open surface (no boundary) whereas the original surface had two boundary components. $\endgroup$ Commented Jun 3, 2017 at 3:36
  • $\begingroup$ @GrumpyParsnip They were open points of course :-) (changed it) $\endgroup$ Commented Jun 3, 2017 at 3:40
  • $\begingroup$ ha!$\phantom{.}$ $\endgroup$ Commented Jun 3, 2017 at 3:47

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