Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 12G (Topics in Analysis)

Moving from $A_1$ to $A_n$, the colour changes an odd number of times because the endpoints have opposite colours: every change toggles the current colour, while every nonchange preserves it. Formally, encode red and green by zero and one; the sum modulo two of adjacent differences is the endpoint difference, which is one.

Count red-green segments modulo two. On $DE$, part (a) says their number is odd; the other two outer sides contain no red-green segment. Every interior segment belongs to two small triangles and therefore contributes zero modulo two, while each boundary segment belongs to one. Thus the sum, over all small triangles, of their numbers of red-green edges is odd.

A triangle with all three colours has exactly one red-green edge. A triangle using only red and green has zero or two, and any other nontrichromatic triangle has zero red-green edges. Hence the parity sum counts precisely the trichromatic triangles modulo two, proving that their number is odd.

Not necessarily, once the subdivision has at least two segments per side. Colour $D,E,F$ red, green, blue respectively and colour every other grid vertex red. No small triangle contains both exceptional vertices $E$ and $F$, and every small triangle therefore uses at most two colours. For the unsubdivided triangle the assertion is of course true.

False. The constant map to $E$ satisfies it: $E\in I$ and also $E\in J\subset J\cup K$.

True. At the vertices the side conditions forceOn each side, $g$ remains in that same side, so the boundary restriction is homotopic within the boundary to the identity and has degree one. If it extended continuously over the filled triangle with image in the boundary circle, that degree-one boundary map would be null-homotopic, which is impossible. Equivalently, a sufficiently fine simplicial approximation would contradict the parity form of Sperner's lemma proved in part (b).