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Assume common Klein bottle immersion to $\mathbb R^3$. Now let $X$ be an image of this immersion without a disc of self intersection (see fig). Proove that $\pi_1(x)$ has presentation $\left<a,b,c \mid aba^{-1}b^{-1}cb^{-1}c^{-1}\right>.$ enter image description here

Any hints how I can see that this is true? Thanks!

Thomas Andrews
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Gleb Chili
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  • I don't understand how $X$ is different from an actual (abstract) Klein bottle. – Arnaud Mortier Jan 22 '18 at 18:12
  • It isn't injective, @ArnaudMortier Specifically, an immersion need not be injective. https://en.wikipedia.org/wiki/Immersion_(mathematics) – Thomas Andrews Jan 22 '18 at 18:20
  • I know what an immersion is, I was just saying that I don't understand what the OP means. Although with $\mathbb{R}^3$ instead of $\mathbb{R}$ it is already much better. – Arnaud Mortier Jan 22 '18 at 18:27
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    @ArnaudMortier Image is homotopic to $S^2 \vee S^1 \vee S^1$: https://math.stackexchange.com/questions/366199/show-a-subspace-formed-by-a-klein-bottle-is-homotopy-equivalent-to-s1-vee-s1 – Gleb Chili Jan 22 '18 at 18:34

1 Answers1

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Start with a square in which you identify the sides in the usual way to represent the Klein bottle, e.g. top and bottom are identified without twist, and left and right are identified with a twist.

Now remove a small disc in the interior of that square. Mark the boundary of that disc to be identified with the twisted side of your square (the one on the right).

That is your immersion, unfolded.

To compute the $\pi_1$ of that, add a small segment joining your inner circle to the boundary (you have to give a name to that segment, that is your third generator). Now what you see is a $2$-dimensional CW-complex with only one $2$-cell. To compute the $\pi_1$ all you need to do is read the boundary of that $2$-cell.

EDIT: the previous version contained a mistake, the correct side with which you identify the boundary of your small disc is the twisted one, because that is the one which has a neighbourhood homeomorphic to a cylinder $\mathbb{R}\times \mathbb{S}^1$ (that you can see on your picture). Here is the picture that you obtain:

enter image description here

Arnaud Mortier
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    Oops, I realise just now that I twisted the top/bottom as well in the picture. One of the triple arrows should be reverted. – Arnaud Mortier Jan 25 '18 at 15:06