User:Lethe/sandbox



stuff

We want to verify the equation
 * $$\tau_A\circ \mathcal{P}(f) = f^*\circ\tau_B$$

where τC: P(C) → 2C is the map which sends any subset of the set C to the characteristic function on that subset, i.e.
 * $$\tau_C(U)=\chi_U,$$

where χU is given by

\chi_U(c) = \begin{cases} 1 & c \in U\\ 0 & c \not\in U \end{cases} $$ for any subset U ⊆ C and any element c ∈ C. To verify the equation, let both sides act on some subset S ⊆ B. We have
 * $$\mathcal{P}(f)(S) = f^{-1}(S)$$

by the definition of the powerset functor, and so
 * $$\tau_A(\mathcal{P}(f)(S)) =\chi_{f^{-1}(S)}.$$

On the right-hand side of the equation, we have
 * $$\tau_B(S) = \chi_S$$

and recall that f* is the pullback by f induced by the contravariant hom-functor; it acts on maps by multiplication on the right:
 * $$f^*(\chi_S)=\chi_S\circ f.$$

So it remains to check the equality
 * $$\chi_{f^{-1}(S)} = \chi_S\circ f.$$

To verify this equation, act both maps in 2A on an arbitrary element a ∈ A.

$$ \chi_{f^{-1}(S)}(a) = \begin{cases} 1 & a\in f^{-1}(S)\\ 0 & a \not\in f^{-1}(S) \end{cases} $$

$$ (\chi_S\circ f)(a)= \begin{cases} 1 & f(a)\in S\\ 0 & f(a) \not\in S \end{cases} $$

Since a ∈ f–1(S) iff f(a) ∈ S, these maps are equal.
 * probably cornbread