The Klein bottle is usually pictured as a subspace of \(\mathbb{R}^3\) like the subspace \(X \subset \mathbb{R}^3\) shown in the first figure at the right. If one wanted a model that could actually function as a bottle, one would delete the open disk bounded by the circle of self-intersection of \(X\), producing a subspace \(Y \subset X\). Show that \(\pi_1(X) \approx \mathbb{Z} * \mathbb{Z}\) and that \(\pi_1(Y)\) has the presentation \(\langle a, b, c \mid aba^{-1}b^{-1}cb^\varepsilon c^{-1} \rangle\) for \(\varepsilon = \pm 1\). (Changing the sign of \(\varepsilon\) gives an isomorphic group, as it happens.) Show also that \(\pi_1(Y)\) is isomorphic to \(\pi_1(\mathbb{R}^3 - Z)\) for \(Z\) the graph shown in the figure. The groups \(\pi_1(X)\) and \(\pi_1(Y)\) are not isomorphic, but this is not easy to prove; see the discussion in Example 1B.13.
Solution
We use the orientation convention giving \(\varepsilon=-1\). Reversing the orientation of the corresponding identified boundary circle gives the convention \(\varepsilon=+1\) without changing the underlying space.
1. The punctured bottle \(Y\) has a CW structure with one \(0\)-cell, three \(1\)-cells \(a,b,c\), and one \(2\)-cell attached along\[ r=aba^{-1}b^{-1}cb^{-1}c^{-1}. \]
Cut the surface in the source figure along arcs from the boundary of the deleted disk to the self-identification curves, so that the remaining \(2\)-cell is a disk. After the cuts, all vertices are identified to one vertex and the three surviving edge classes are \(a,b,c\). Traversing the boundary of the cut-open disk once gives, in order, \[ a,\ b,\ a^{-1},\ b^{-1},\ c,\ b^{-1},\ c^{-1}. \] Thus the attaching word is exactly \(r\). This is the standard one-disk fundamental polygon obtained by opening the bottle picture along the deleted self-intersection disk.
2. Therefore\[ \boxed{ \pi_1(Y) \cong \left\langle a,b,c\ \middle|\ aba^{-1}b^{-1}cb^{-1}c^{-1}=1 \right\rangle . } \]
The \(1\)-skeleton in 1 is \[ S^1\vee S^1\vee S^1, \] so its fundamental group is the free group \(F(a,b,c)\). Attaching the single \(2\)-cell quotients by the normal closure of its attaching word \(r\). This is precisely the displayed presentation by van Kampen.
3. With the opposite orientation convention on the indicated boundary identification, the same computation gives\[ \pi_1(Y) \cong \left\langle a,b,c\ \middle|\ aba^{-1}b^{-1}cb c^{-1}=1 \right\rangle . \]
Only the orientation in which that copy of the \(b\)-edge is traversed changes. Since this is merely a change of orientation in the same CW decomposition of the same space \(Y\), the two presentations present isomorphic groups. Thus the answer can be written uniformly as \[ \left\langle a,b,c\ \middle|\ aba^{-1}b^{-1}cb^{\varepsilon}c^{-1}=1 \right\rangle, \qquad \varepsilon=\pm1. \]
4. Reattaching the deleted disk to \(Y\) to recover the immersed Klein-bottle model \(X\) kills the generator \(b\) in the presentation of
2.
In the cut-open model of 1, the boundary of the disk deleted at the circle of self-intersection is represented by the \(b\)-loop. Passing from \(Y\) back to \(X\) attaches a \(2\)-cell along this boundary loop, so van Kampen adds the relation \[ b=1. \]
5. Hence\[ \boxed{\pi_1(X)\cong\mathbb Z*\mathbb Z.} \]
Adding \(b=1\) to the presentation in 2 makes the original relator trivial: \[ aba^{-1}b^{-1}cb^{-1}c^{-1} \longmapsto aa^{-1}cc^{-1}=1. \] Therefore \[ \pi_1(X) \cong \langle a,c\mid\ \rangle \cong F(a,c) \cong\mathbb Z*\mathbb Z. \]
6. The graph complement \(\mathbb R^3\setminus Z\) has the same fundamental group as \(Y\).
Use the graph \(Z\) exactly as embedded in the rightmost source figure. Take a sufficiently small regular neighborhood of the pictured graph and use the transverse cross-sections indicated by the drawing. In every cross-section, the complement of the central graph point radially retracts onto the corresponding arc of the bottle surface \(Y\). These radial retractions agree along the edge and vertex cross-sections, so over a large ball containing \(Z\) they give a deformation onto the pictured copy of \(Y\) together with the outer boundary sphere.
Outside that ball, radial projection retracts the unbounded complement onto the same outer sphere. Consequently \[ \mathbb R^3\setminus Z\simeq Y\vee S^2. \] Since \(S^2\) is simply connected, van Kampen gives \[ \pi_1(\mathbb R^3\setminus Z) \cong \pi_1(Y)*\pi_1(S^2) \cong \pi_1(Y). \]
7. Thus all requested groups are\[ \pi_1(X)\cong\mathbb Z*\mathbb Z, \qquad \pi_1(Y) \cong \pi_1(\mathbb R^3\setminus Z) \cong \left\langle a,b,c\mid aba^{-1}b^{-1}cb^{\varepsilon}c^{-1}\right\rangle . \]
Combine <1>2–<1>6.