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I saw the same question posted here. However, by the answer of Ka Ho, I get a torus instead of a Klein bottle. Since they are not homotopy equivalent, this would mean that the quotient of $\Delta^3$ in question cannot deformation retract to a Klein bottle right?

Edit: In the same exercise, how to get a 2-sphere from identifications of edges and deformation retract?

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The idea is to collapse the face $[v_0,v_1,v_2]$ so that the edge $[v_1,v_2]$ identifies with $[v_1,v_0]+[v_0,v_2]$. From here it should be relatively easy to see how to flatten the rest of $\Delta^3$ and one is left with a square with edges oriented edges, give them names $[v_0,v_1]=a,\, [v_0,v_2]=b$ such that $a^2b^2$ is a nullhomotopic. If one does not recognize this as the Klein bottle, go to pg. 51 there in Hatcher and this complex is illustrated.

The take away from this is to find a square on $\Delta^3$ and identify these edges and leave the faces and other edges undisturbed. For the sphere, one could make identifications $[v_0,v_1] \sim -[v_1,v_3]$ and $[v_0,v_2] \sim -[v_2,v_3]$.

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  • $\begingroup$ Thanks. Now I see where I got it wrong. I always thought that we were identifying opposing edges. $\endgroup$ – baby bunny Sep 15 '17 at 8:48

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