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.

$\tau _{C}(U)=\chi _{U},$

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

$\tau _{B}(S)=\chi _{S}$

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 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