# Show that the $\Delta$-complex obtained from $\Delta^3$ by performing edge identifications deformation retracts onto a Klein bottle.

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?

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.
flatten the tetrahedron and draw it in the plane (triangle with a vertex inside and edges going out to the vertices of the triangle). if you cut it up a little, you're looking at the standard "rectangle-with-sides-identified" picture of the klein bottle. sorry for the terrible picture, mspaint hasnt changed since 3.x as far as i can tell...
edit: after "smooshing" the tetrahedron (set it on the table and press down), you have the first triangle. deforming away the black triangle gives the second picture (ignoring all the letters). we have $$a=[v_0,v_1]=[v_1,v_3]$$, $$b=[v_0,v_2]=[v_2,v_3]$$, and i'm introducing new edges $$c$$ and $$d$$. cutting the second triangle into two rectangles (both with edge labels $$a,b,c,d$$), then regluing along $$a$$ gives you a rectangle. this is the "standard" klein bottle.
the left two rectangles are what you get by cutting the second triangle along $$c,d$$. the right two are supposed to indicate regluing along $$a$$, but there's a mistake in the labeling. (sorry i don't want to redraw a picture, i answered this like 5 years ago.)