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I am going through some exercises in Hatcher's Algebraic Topology. You have a $\Delta$-complex obtained from $\Delta^3$ (a tetrahedron) and perform edge identifications $[v_0,v_1]\sim[v_1,v_3]$ and $[v_0,v_2]\sim[v_2,v_3]$. How can you show that this deformation retracts onto a Klein bottle?

Batominovski
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09867
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2 Answers2

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Flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle); i.e. smoosh the 3-cell. If you smoosh and cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the Klein bottle.

enter image description here

yoyo
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  • I remember from doing this exercise that the rectangle-with-sides-identified isn't quite the standard one (by which I mean pairs of opposite sides identified, one with a twist). The edge orientations that are specified by the delta-complex structure mean that you end up with something that needs a little cutting and gluing to see that it is your friendly ordinary klein bottle. – NKS Jan 18 '12 at 16:50
  • Bit confused about how to go about the squishing of it. Whenever I try it doesn't get to the Klein bottle square. – 09867 Jan 18 '12 at 16:52
  • @NKS yes you do have to cut it up, my bad – yoyo Jan 18 '12 at 19:12
  • @yoyo May I ask what does the black shaded part represent? Also why is the vertex 0 in the center in the first diagram but not in the center for the larger second diagram? Thanks! – yoyostein Jun 01 '16 at 09:28
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The 3-simplex obviously deformation retracts onto the union of the surfaces obtained by $[v_0,v_1,v_3]$ and $[v_0,v_2,v_3]$. Note that the continuous image of a deformation retract, where the map identifies the points in the retract only, is still a deformation retract.

user458870
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