Math 4 Wisdom. "Mathematics for Wisdom" by Andrius Kulikauskas.

Classify adjunctions, Examples of adjunction, Induction restriction adjunction, Adjunction in statistics, Category theory, Limits vs colimits, Equivalence, Sameness, Definition of adjunction, Nonexistence of adjunction

Given a functor, provide an algorithm to show that it does not have a left adjoint functor or a right adjoint functor.

Nonexistence of adjunction

The inclusion of Field in CRing has no left adjoint because it would carry Z to an initial field, which does not exist. How might an initial field relate to the field with one element?

Use the definition of adjunction in terms of the universal mapping property. Rephrase it in the negative.

Example: Inverse image functor is not the left adjoint of the direct image functor.

Given {$f:A\rightarrow B$} consider the preorders {$PA$} and {$PB$} of the subsets of {$A$} and {$B$}, respectively. The inverse image functor {$f^{-1}:PB\rightarrow PA$} is defined by {$f^{-1}(B')=\{ a\in A | f(a)\in B' \}$}. The direct image functor {$f_*:PA\rightarrow PB$} is defined by {$f_*(A')=\{ b\in B | f(a)\in B' \}$}.

{$\textrm{Hom}_{PB}(f^{-1}(B'),A')$} is not isomorphic to {$\textrm{Hom}_{PA}(B',f_*(A'))$}

As long as {$f$} is not surjective, we can define {$A'$}, {$B'$} so that there exists {$b\in B'$}, {$b\notin f_*(A)$}.

Then we may have {$f^{-1}(B')\leq A'$} but we have {$b\notin f_*(A')$} and so {$B' \nleqslant f_*(A')$}.

Alternatively, if {$f$} is not injective, then we can define {$a'\in A'$}, {$a''\in A$}, {$a''\notin A'$} such that {$f(a')=f(a'')\in B'$}.

Then we may have {$B'\leq f_*(A')$} and yet {$a''\notin A'$} and so {$f^{-1}(B') \nleqslant A'$}.

However, if {$f$} is both surjective and injective, then the inverse image functor is the left adjoint of the direct image functor. And, as always, it is also the right adjoint of the direct image functor.