Pullback preserving Set-functors

Pullback preserving Set-functors

[Tobias Schroeder]

I'm interested in proofs of or counterexamples for
the following conjectures:

Conjecture 1:
Any Set-endofunctor that preserves
kernels (i.e. pullbacks of a mapping with itself)
preserves pullbacks.

Conjecture 2:
Any Set-endofunctor that preserves
kernels *and inverse images* (i.e. pullbacks
where one of the mapping is injective)
preserves pullbacks.

Conjecture 3:
Same as Conj. 2 with *and inverse images*
replaced by *and equalizers*.

Conjecture 4:
Same as Conj. 1-3, but concerning *weak*
preservation.

Conjectur 5:
Same as 1-4, but for Set-endofunctors
that are subfunctors of a pullback
preserving functor.

I tried to prove these facts in several ways
but was not able to do it or to find a
counterexample (the answer to this questions
is of some relevance for my work on coalgebras)
... and it looks quite easy, doesn't it?

Can somebody help me in this?

Thank you very much in advance

Tobias Schröder
--------------------------------------------------------------
Tobias Schröder
FB Mathematik und Informatik
Philipps-Universität Marburg
WWW: http://www.mathematik.uni-marburg.de/~tschroed
email: tschroed@mathematik.uni-marburg.de

re: Pullback preserving Set-functors

[Peter Freyd]

Tobias Schroeder asks:


    Conjecture 1:
    Any Set-endofunctor that preserves kernel pairs
    [He called them "kernels"]
    preserves pullbacks.

Counterexample: the functor that sends the empty set to the empty set
and everything else to a fixed one-element set. (Note that this is
also a counterexample for the conjecture-5 modification.)


    Conjecture 2:
    Any Set-endofunctor that preserves n
    kernel pairs *and inverse images* (i.e. pullbacks
    where one of the mapping is injective)
    preserves pullbacks.

Proof: Given such a functor, T:A --> B, where  A  and  B  are 
sufficiently nice categories, lift it to  T:A --> B/T1  to obtain a
functor that not only preserves kernel-pairs and inverse images but
the terminator. It suffices to prove that this lifted  T  preserves
pullbacks (using the fact that the forgetful functor  B/T1 --> B 
preserves pullbacks). hence it suffices to consider the conjecture
assuming that the functor preserves also the terminator. A well-known
argument then reduces the question to the preservation of binary 
products. (One may construct equalizers using inverse images and
products and from there to arbitrary pullbacks is ancient. See 1.43 in
Cats and Allegators.) What we will use is that the functor preserves 
inverse images and iterated products. Given sets  X  and  Y  we know 
that the product diagram:

             (X+Y)x(X+Y)

                /   \         [add your own downwards arrowheads]

             X+Y     X+Y

is carried to a product diagram

           T((X+Y)x(X+Y))

                /   \  

           T(X+Y)   T(X+Y).

Using three inverse-image diagrams obtain:

                XxY

               /   \

           Xx(X+Y) (X+Y)xY

           /   \   /   \

         X  (X+Y)x(X+Y)  Y

           \   /   \   /       

             X+Y     X+Y

Now apply  T:

              T(XxY)

              /     \

       T(Xx(X+Y)) T((X+Y)xY)

         /     \   /     \

       TX  T((X+Y)x(X+Y)) TY

         \     /   \     /       

          T(X+Y)  T(X+Y)

As already observed, the very bottom  / \  is a product diagram. Each
rombus is an inverse-image pullback, hence preserved. Thus the top
 / \
/   \  a product diagram.



    Conjecture 3:
    Same as Conj. 2 with *and inverse images*
    replaced by *and equalizers*.

Inverse images of regular subobjects (that is, those that appear as
equalizers) are regular: the equalizer of  x,y  under a map  f  is the
equalizer of  xf,yf.  Hence if every subobject is regular than
preservation of equalizers implies preservation of inverse images.