Consider a Mobius strip, colored symmetrically: Yellow-Green-Blue-Green. (Actually, there are two shades of blue and accents of orange, purple, and magenta, but these extra colors do not contribute to analysis.) If the Mobius strip were non-self-intersecting, a 4*PI cycle could be written as a word, Y1G1B1g1Y2G2B2g2. Instead require that B2 is bisected by the inner Y edge. The path through B2 is obstructed, and the 4*Pi cycle can not be completed, see for example: https://0x0.st/zpTf.JPG https://0x0.st/zpT9.JPG https://0x0.st/zpTp.JPG https://0x0.st/zpTL.JPG Fix the topology by allowing that Y1 and Y2 connect directly to the left and right parts of B2. PROBLEM: How many homology classes does the YGB sculpture have? ANSWER: Three paths are valid from Y1 to Y2: Y1G1B1g1Y2, Y1B2G2Y2, and Y1g2B2Y2. An entire cycle involves one half path plus the reverse of another. Three choose two equals three. Thus the geometry has three homology classes, two of length 7, one of length 6. Explicitly they are: Y1G1B1g1Y2G2B2, Y1G1B1g1Y2B2g2, and Y1B2G2Y2B2g2. A Mobius strip has only one homology class, so this YGB Surface must be something else entirely. Is it known already? If yes, does it also have a vanity name after a famous mathematician? Cheers, Brad