Neil Ghani's question

Neil Ghani's question

[Michael Barr]

A week or two ago, Neil Ghani asked about natural transformations
between set-valued functors (I think they were set-valued, but anyway
that is what my answer refers to and is probably true for any reasonably
complete codomain category although a different argument would be
required), say a: F ---> G, such that for any arrow f: A ---> B of the
domain category, the square
                      aA
                FA --------> GA
                |            |
                |            |
                |Ff          |Gf
                |            |
                |            |
                v     aB     v
                FB --------> GB
 is a pullback.  At the time, I sent Neil a private reply, but it
bounced for some reason.  (I said that that I thought that this
condition was reasonable only when restricted to monic f and then such
an a is called an elementary embedding.)  Then a couple of people
answered that it was called a cartesian arrow and I didn't try to resend
my answer.  Well, there is a simpler answer.  In that generality, such
an a is called a natural equivalence.  In other words, non-trivial
examples do not exist.

To see this, it is useful to translate it, using Yoneda, into the
following form.  As usual, I will say that of two classes E and M of
arrows in a category, E _|_ M (E is orthogonal to M) if in any diagram
                  e
              A ----> B
              |       |
              |       |
              |       |
              v   m   v
              C ----> D
 with e in E and m in M, there is a unique arrow B ---> C (called a
diagonal fill-in) making both triangles commute.  Let us denote by h^A
the covariant functor represented by A and for f:  A ---> B, denote by
h^f, the induced natural transformation h^B ---> h^A.  Let E be the
class of all h^f.  Then a is cartesian iff E _|_ {a}.

Now suppose a is cartesian.  First I show that a is monic (that is
injective).  If not, there is an object A and two different arrows u, v:
h^A ---> F such that aA(u) = aA(v).  Let E be the equalizer of u and v
and let h^B ---> E be any arrow.  Then the square

                 B        A
                h -----> h
                |        |
                |        |aA(u)=aA(v)
                |        |
                v   a    v
                F -----> G
 has two diagonal fill-ins, u and v.  Here the arrow h^B ---> h^A is the
composite h^B ---> E ---> h^A and the arrow h^B ---> F is the composite
h^B ---> E ---> h^A ---> F, the latter via u or v.  In a similar way, we
can show that a is surjective.  In fact, given u: h^A ---> G, let E be a
pullback of a and u and let h^B ---> E be arbitrary.  Then we have a
commutative square

                B         A
               h  -----> h
               |         |
               |         |u
               |         |
               v    a    v
               F ------> G
 whose diagonal fill-in gives a lifting of u.

It therefore seems appropriate to restrict the question to certain
classes of arrows A ---> B, for example monics.  Here are a couple of
examples.  If g:  C --->> D is a regular (or just strict) epimorphism
between objects of the domain category, then for E the class of h^f for
all monic f, we have E _|_ {h^g}.  Similarly if E is the class of h^f
for all strict monic f and g is any epimorphism, E _|_ {h^g}.

Re: Neil Ghani's question

[Dr. P.T. Johnstone]

Mike Barr's answer to Neil Ghani's question is very pretty, but
unfortunately wrong. There are many examples of cartesian natural
transformations (e.g. between functors Set --> Set) which are neither
epic nor monic: for example, the natural transformation
(-) x A --> (-) x B induced by an arbitrary map A --> B.

The mistake in Mike's proof occurs when he says

> Let E be the equalizer of u and v
> and let h^B ---> E be any arrow.

The trouble is that E could be the zero functor, so that there may
not be any arrows h^B --> E.

Peter Johnstone

Re: Neil Ghani's question

[Max Kelly]

I refer to Michael Barr's comments on Neil Ghani's question on cartesian
natural transformations. These have been much studied, especially in 
computer science conrexts, and Michael admits he had seen the replies of
Peter Johnstone and myself, in which we independently give precise
references to our separate contributions to the study of monads whose
multiplication and unit are cartesian natural transformations; such as
the monad whose algebras are pointed sets, the multiplication for which
is the cartesian natural transformation whose A-component is A+1+1 --> A+1.

Accordingly I thought it odd that Michael, in the face of this, trusted
his proof that they exist only trivially. Of course, as Peter Johnstone
said, Michael's E is usually empty. Ironically, Michael is tha author of
a famous and striking paper on the point of the empty set, which inter
alia points out earlier errors of this kind on the part of others.

Max Kelly.