# Quotient of a triangle

This is an exercise in Hatcher's Algebraic topology:

What familiar space is the quotient $\Delta$-complex of a $2$-simplex $[v_0,v_1,v_2]$ obtained by identifying the edges $[v_0,v_1]$ and $[v_1,v_2]$, preserving the ordering of the edges?

The equivalence relation is thus generated by $tv_0 +(1-t)v_1 \sim tv_1+(1-t)v_2$. It is no problem to calculate the homology groups which are the same as for the circle $S^1$. The quotient could thus be homotopy equivalent to $S^1$. Nevertheless, I don't SEE this familiar space (or, more precisely, a homeomorphic copy in $\mathbb R^3$).

Let $v_3$ be a point on $[v_0,v_2]$ and cut the $2$-simplex across the line segment from $v_1$ to $v_3$. This cuts it into two $2$-simplices, one with vertices $[v_0,v_1,v_3]$ and another with vertices $[v_1',v_2,v_3']$. The edge $[v_0,v_1]$ is identified with $[v_1,v_2]$, and the edge $[v_1,v_3]$ is identified with $[v_1',v_3']$. Now the trick is to glue these $2$-simplices back together along $[v_0,v_1]$ and $[v_1,v_2]$ instead of along $[v_1,v_3]$. When you perform this gluing, you should get something familiar. The answer is hidden below.
This gives a square with vertices $v_1'$, $v_3'$, $v_1$, and $v_3$, with $[v_1',v_3']$ identified with $[v_1,v_3]$. This is exactly the standard presentation of a Möbius strip as a quotient of a square!