Re: making a cone universal in a faithful way

[Lutz Schroeder <Lutz.Schroeder@dfki.de>]

Dear Dimitri,

> For example, if I is the empty category, the question becomes "when can
> you make c an initial object in a faithful way?". If I is the final
> category, then the cocone alpha amounts to a morphism f : F(*) -> c and the
> question becomes "when can you make f an isomorphism in a faithful way?".
>
> There are two obvious necessary conditions.
> 1) Let f,g : c -> d be two morphisms of C. If f alpha_i = g alpha_i for
> every i in I, then we should have f = g.L
> 2) Let f : a -> F(i) and g : a -> F(j) be two morphisms of C such that
> alpha_i f = alpha_j g. Then for every cocone beta : F => d, we should
> have beta_i f = beta_j g.
>
> In the case of the empty category, the first condition means that for
> every object d there is at most one arrow c -> d and the second condition
> is void. In the case of the final category, the first condition means
> that f is an epi and the second that f is a mono. It is not hard to prove
> that in both cases, theses conditions are sufficient.

While I'm willing to believe the case of initial objects, the statement
is wrong already for the case of isomorphisms. Let's call f a potential
isomorphism if there exists a faithful functor (equivalently an
embedding) that makes f an isomorphism. Then one has the following
property of potential isomorphisms (from my 1999 thesis; in German, I'm
afraid):

Lemma: Let s be a potential isomorphism, and let f,g,h,j,l,p be
morphisms such that

	fs = sg
	hg = ks
	fl = sp.

Then kl = hp.

Proof: In an extended category where s has a two-sided inverse s^{-1},
we have gs^{-1} = s^{-1}f from fs = sg, and hence

	kl = kss^{-1}l = hgs^{-1}l = hs^{-1}fl = hs^{-1}sp = hp    []

The property of the lemma is not implied by s being both epi and mono
(i.e. a bimorphism). It is comparatatively easy to prove this using
contrived examples, such as the following.

Let the category A consist of objects A, B, C, D, and families of morphisms

	g_i: A -> A
	p_i: B -> A
	s_k: A -> C
	h_i: A -> D
	l_i: B -> C
	f_i: C -> C
	k_i: B -> D
	q_i,r: B -> D

indexed over i>=0, k>=1, where we identify f_0 and g_0 with the
respective identities. Composition is by addition of indices, with the
single exception

	h_0p_0 = r.

(This satisfies the associative law, since the exceptional case r occurs
only in trivial cases of the law -- it cannot be pre- or postcomposed
with a nontrivial morphism, and its two factors do not have proper
factorisations.) Then s_1 is a bimorphism but violates the property of
the above lemma, and hence is not a potential isomorphism:

	f_1s_1 = s_1g_1, h_0g_1 = k_0s_1, and f_1l_0 = s_1p_0,

but

	k_0l_0 = q_0 \neq r = h_0p_0.


Best regards,

Lutz

-- 
--------------------------------------
PD Dr. Lutz Schro"der
Senior Researcher
DFKI Bremen
Safe and Secure Cognitive Systems
Cartesium, Enrique-Schmidt-Str. 5
D-28359 Bremen

phone: (+49) 421-218-64216
Fax:   (+49) 421-218-9864216
mail: Lutz.Schroeder@dfki.de
www.dfki.de/sks/staff/lschrode
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