Take a 2-simplex $[v_0, v_1, v_2]$ and form the quotient space given by identifying the faces (edges) $[v_0, v_1]$ and $[v_1, v_2]$ preserving the ordering of vertices.

The resulting quotient space will be a compact surface with boundary, and is homeomorphic to a Mobius strip. We can see this with the classification of compact surfaces (with boundary):

After identifying the edges, there is one vertex, two edges, and one face, giving $\chi=0$, then observing that the the surface is non-orientable (because we have a pair of edges identified in the 'same direction' around the polygon), so the surface is either $K^2$ or $RP^2 - D$ (the Mobius strip), and since there is at least one boundary component (along the unidentified edge, for instance), it must be $RP^2 - D$.