Subject: categories: Neil Ghani's question

From: Michael Barr <[email-masked]>
Date: Sat, 4 Apr 1998 09:07:35 -0500 (EST)

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}.

---

From: "Dr. P.T. Johnstone" <[email-masked]>
Date: Sat, 4 Apr 1998 15:57:18 +0100 (BST)

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