From: Michael Barr <barr@triples.math.mcgill.ca>
To: categories@mta.ca
Subject: Neil Ghani's question
Date: Sat, 4 Apr 1998 09:07:35 -0500 (EST) [thread overview]
Message-ID: <Pine.LNX.3.95.980404090557.9640B-100000@triples.math.mcgill.ca> (raw)
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}.
next reply other threads:[~1998-04-04 14:07 UTC|newest]
Thread overview: 3+ messages / expand[flat|nested] mbox.gz Atom feed top
1998-04-04 14:07 Michael Barr [this message]
1998-04-04 14:57 ` Dr. P.T. Johnstone
1998-04-06 2:06 Max Kelly
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