Let
Grp denote the
category of groups and
Grpiso denote
Grp restricted to contain only those
morphisms which are
isomorphisms.
Now let
F:Grpiso→Grp be given by
F(A)F(f)=Aut(A)=(φ↦fφf−1)on objectson morphisms
Let’s check first that this definition makes any sense. On
objects,
F takes a
group A∈Grpiso and produces
Aut(A), which is indeed in
Grp. So far so good. On
morphisms,
F takes a
group homomorphism f:A→B and is supposed to produce a
group homomorphism F(f):Aut(A)→Aut(B). This
group homomorphism takes an element
φ∈Aut(A) and produces
fφf−1∈Aut(B).
Perhaps the most subtle aspect of this definition is how we define
F(f), so let’s see it diagramatically:
Given
f:A→B an
isomorphism, we need that
F(f) be
Aut(A)→Aut(B). That is,
F(f) must take an
isomorphism φ:A≅A and produce an
isomorphism F(f)(φ):B≅B. It is able to do this because
f is itself an
isomorphism, meaning that it has an inverse. The
morphism F(f)(φ) uses
f−1 to go from
B to
A, then uses
φ to automorph in
A, then uses
f to go back to
B, in all producing an
automorphism B≅B. The result
F(f)(φ)=fφf−1 is an
isomorphism because it has inverse
fφ−1f−1
With
F defined, we still need see that it is actually a functor. We’ll just bash this out:
Does
F preserve identities? Yes:
F(1A)=(φ↦1Aφ1A−1)=(φ↦φ)=1Aut(A)=1F(A)
Does
F respect composition? Yes:
F(g∘f)=φ↦(g∘f)∘φ∘(g∘f)−1=φ↦g∘f∘φ∘f−1∘g−1=φ↦g∘F(f)(φ)∘g−1=φF(g)(F(f)(φ))=F(g)∘F(f)
Given an isomorphism, we need that be . That is, must take an isomorphism and produce an isomorphism . It is able to do this because is itself an isomorphism, meaning that it has an inverse. The morphism uses to go from to , then uses to automorph in , then uses to go back to , in all producing an automorphism . The result is an isomorphism because it has inverse
With defined, we still need see that it is actually a functor. We’ll just bash this out: