User:Lethe/sandbox

stuff

We want to verify the equation

$\tau_A\circ \mathcal{P}(f) = f^*\circ\tau_B$

where τ_C: P(C) → 2^C is the map which sends any subset of the set C to the characteristic function on that subset, i.e.

τ C (U) = χ U,

where χ_U 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

τ B (S) = χ 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 2^A 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