2
$\begingroup$

In category theory a morphism is constant IIF it is absorbing (for left composition).
That is a morphism $k$ from $k:A\rightarrow B$ is constant if an only if for any two parrallel (same domain and same codomain) morphisms $f$ and $g$ of codomain $A$ we have $k(f) = k(g)$.

Now $c$ is co-constant IIF it is constant in the opposite category (equationally $f(c) = g(c)$ for any $f$ ...) .
Semantically (in Set at least) constness is rather clear , that is a non mathematician understands the idea of "unchanging", the question(s) is:

Q1: For constness are there other semantic appearing in something else than $Set$.

Q2 :What does it mean (semantically) to be coconstant in $Set$ and elsewhere ?

Note: I guess there are several (categorically prototypical) answers in both questions.

Remark : I thought of something like an unchanging measure of noise (constant) versus an intrinsically flat noise (coconstant). But this is a rather foggy intuition, moreover I cannot categorify it properly.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Remark : The definition of constant is the one from nLab not that of Lawvere (factoring through the terminal object). $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Mar 5, 2011 at 14:17

1 Answer 1

Reset to default
2
$\begingroup$

Edit: This is not quite right -- the two definitions of constantness are not equivalent even in the category of sets, which means that the representability argument doesn't quite work. See the nlab page for details, but note that they are equivalent if the hom sets $C(X,A)$ are inhabited for every X, or equivalently if A admits a global section.

This doesn't answer all of your questions, but...

A morphism is constant in your sense iff it is representably constant in (what you say is) Lawvere's sense, that is if $k_* \colon C(X,A) \to C(X,B)$ factors through the terminal object in Set. So if C itself has a terminal object then the two are equivalent.

A morphism c is co-constant iff $c^* \colon C(B,Y) \to C(A,Y)$ is constant, and it's easy to see that if C has an initial object then it's equivalent to ask that c factor through it. But in Set the initial object $\emptyset$ is strict, meaning that any morphism into $\emptyset$ is an isomorphism. So there are no non-trivial co-constant morphisms in Set, or indeed in any category with a strict initial object (such as (by a result of Joyal) any cartesian closed category with initial object, like a topos).

I can't tell off the top of my head whether your definition would be useful elsewhere, but the above does put some restrictions on what sorts of categories non-trivial co-constant morphisms might turn up in.

Improve this answer
$\endgroup$
7
  • $\begingroup$ @Finn : The Lawvere definition implies the nlab definition, but i cannot see the reverse implication in a category with a terminal object. $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Mar 5, 2011 at 21:52
  • $\begingroup$ If each function $f \mapsto kf$ factors through the terminal set, then the natural transformation $k_* \colon C(-,A) \to C(-,B)$ factors through the constant functor at the terminal set. If C has a terminal object T then it represents that functor, so by the full faithfulness of the Yoneda embedding $k = ! \circ b$ for some $b \colon T \to B$, where $! \colon A \to T$. $\endgroup$
    – Finn Lawler
    Commented Mar 5, 2011 at 23:15
  • $\begingroup$ @Finn : Sorry I still do not see how this show that nlab definition ( on morphisms not functors) implies Lawvere ( assuming only a terminal object) ? $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Mar 13, 2011 at 1:11
  • $\begingroup$ Do you understand why it is that if a morphism k is nLab-constant in C then the natural transformation $C(-,k)$ is Lawvere-constant in $[C^{\mathrm{op}}, \mathrm{Set}]$? $\endgroup$
    – Finn Lawler
    Commented Mar 13, 2011 at 16:58
  • $\begingroup$ @Finn :I found a counter example (C.E.): Let $C$ has two objects $A$ and $B$, an arrow $f$ from $A$ to $B$ and the two identities arrows then $f$ is nLab constant (vacuously) but NOT Lawvere. Another C.E. is $A={x,y}$ ,B={u,v} ,C={u',v'} , T={t}$ as morphisms add all applications to $T$ (the terminal element). Add morphisms $f,g,h$ with $f(u)=u',f(v)=v'$ , $g(x)=u ,g(y)=v$ , $h(x)=v ,h(y)=u$. Now clearly $f(g)=f(h)$ : $f$ is nLab-constant but not Lawvere. More generally if $g$ and $h$ have the same $B$ fibers it gives a C.E.. The idea is nLab means "absorbing" , Lawvere is "one valued". $\endgroup$
    – Jérôme JEAN-CHARLES
    Commented Apr 20, 2011 at 11:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .