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Cake day: June 12th, 2023

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  • If the host were to pick a door randomly, there would be 6 equally likely possibilities.

    First, you pick either Goat 1, Goat 2, or Car. In the first case (1/3 chance), the host picks either Goat 2 or Car. In the second (1/3 chance), the host picks either Goat 1 or Car. In the last (1/3 chance), the host picks either Goat 1 or Goat 2.

    Out of these 6 possibilities, two of them result in the host revealing a car, which would end the game early. Eliminating those two possibilities, so the host always reveals a goat, leaves 4 possibilities. This is the “new information” that is used by the host.

    In the first case (1/3 chance), switching gives you the car. In the second case (1/3 chance), switching gives you the car. In the last case (1/3 chance), switching gives you a goat.





  • 🍕(–, B) : C -> Set denotes the contravariant hom functor, normally written Hom(–, B). In this case, C is a category, and B is a fixed object in that category. The – can be replaced by either an object or morphism of C, and that defines a map from C to Set.

    For any given object X in C, the hom-set Hom(X, C) is the set of morphisms X -> B in C. For a morphism f : X -> Y in C, the Set morphism Hom(f, B) : Hom(Y, B) -> Hom(X, B) is defined by sending each g : Y -> B to gf : X -> B. This is the mapping C -> Set defined by Hom(–, C), and it’s a (contravariant) functor because it respects composition: if h : X -> Y and f : Y -> Z then fh : X -> Z and Hom(fh, C) = Hom(h, C)Hom(f, C) sends g : Z -> B to gfh : X -> B.

    P^(n)® AKA RP^n is the n-dimensional real projective space.

    The caveat “phi is a morphism” is probably just to clarify that we’re talking about “all morphisms X -> Y [in a given category]” and not simply all functions or something.

    For more context, the derived functor of Hom(–, B) is called the Ext functor, and the exactness of that sequence (if the typo were fixed) is the statement of the universal coefficient theorem (for cohomology): https://en.wikipedia.org/wiki/Universal_coefficient_theorem The solution to this problem is the “Example: mod 2 cohomology of the real projective space” on that page. It’s (Z/2Z)[x] / <x^(n+1)> or 🍔[x]/<x^(n+1)>, i.e. the ring of polynomials of degree n or less with coefficients in 🍔 = Z/2Z, meaning coefficients of 0 or 1.