Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A constant morphism $f \in \mathrm{Hom}(X,Y)$ is a morphism such that for any object $Z$ and any morphisms $g,h \in \mathrm{Hom}(Z,X)$, $f \circ g = f \circ h$. This is very easy to grasp and one can think of many examples. For instance, in the category of continuous real-valued functions, any map $f(x) = k$ where $k$ is a real number proves to be a constant morphism.

On the other hand, a morphism $f \in \mathrm{Hom}(X,Y)$ is coconstant if for any object $Z$ and morphisms $g,h \in \mathrm{Hom}(Y, Z)$ we have $g \circ f = h \circ f$. In this case, the only real-valued function that I can think of that satisfies this is the $0$-map.

What are some good, nontrivial examples of coconstant functions that can provide intuition for understanding this concept? That is, other than the obvious use of this device to provide duality for the constant morphism, does it have any deeper meaning/significance? I understand what the definition means but don't really see its use beyond the simple examples given.

share|improve this question
    
Well, formally, if you understand constant morphisms you surely understand coconstant morphisms, since they are just constant morphisms in the opposite category :P –  Bruno Stonek Jun 14 '12 at 14:46
    
Something isn't quite right in your definition of a constant morphism. I would suggest $g,h \in \text{Hom}(Z,X)$ or something similar if you read your compositions in the other direction. –  Nate Iverson Jun 14 '12 at 15:01
    
@NateIverson Thanks for pointing out the typo; it has been fixed –  ItsNotObvious Jun 14 '12 at 15:26

3 Answers 3

up vote 5 down vote accepted

Suppose that your category has an initial object $I$.
Then any composition $f:X\stackrel {u}{\to} I \stackrel {v}{\to} Y$ is co-constant .
Indeed if you have two morphisms $g,h:Y\to Z$ and if $i:I\to Z$ is the unique morphism, then:
$$g\circ f=g\circ v\circ u=i\circ u=h\circ v\circ u=h\circ f$$ In other words, any morphism factorizable through $I$ is co-constant.
In particular any morphism with source $I$ is co-constant.

Example
In the category of commutative rings $ \mathbb Z$ is an initial object.
So, given a ring $A$, the morphism $f:\mathbb Z[X]\to A:P(X)\mapsto P(17)\cdot 1_A$ is co-constant since it is the composition $\mathbb Z[X]\stackrel {eval_{17}}\to \mathbb Z\to A:P(X)\mapsto P(17)\mapsto P(17)\cdot 1_A$

share|improve this answer
    
Ah, OK, that's an interesting example and illuminates why conconstance is worth considering. –  ItsNotObvious Jun 14 '12 at 18:08

This is the intuition behind it: In some contexts you can talk about constant functions (eg. set theory, linear algebra, topology, ect). Smart categorists generalized the idea to all categories with: constant morphism

In some other contexts you can talk about functions going to 0 or going to some neutral element (eg. linear algebra, group/ring theory, ect). Smart categorists generalized the idea to all categories with: coconstant morphism.

using category theory you discover that these ideas (constant function vs. function to 0) are dually related. Something that is not too obvious.

Also:

If a category has a terminal object, then any morphism going to the terminal object is a constant morphism and , guess what, categorists called the generic terminal object 1 (the archetipal constant).

If a category has an initial object, then any morphism coming from the initial object is a coconstant morphism and , guess what, categorists called the generic initial object 0 (the archetipal neutral element).

So we have a happy circumstance of nice ideas supported by suggestive notation.

See also Zero morphism for more examples

share|improve this answer

Take a function $f:\{k\} \to Y$ for any real number $k$ and $Y \subseteq \mathbb{R}$

share|improve this answer
    
hmm.. this doesn't work either because $g(f(k))$ will not be equal to $h(f(k))$ unless we force it. Perhaps if you require a portion of linearity for all your morphisms $f(0)=g(0)=h(0)=0$. My answer probably should be deleted. No time now, maybe after class... –  Nate Iverson Jun 14 '12 at 15:53
    
The identity homomorphism should meet the criteria but this is trivial.. –  Nate Iverson Jun 14 '12 at 15:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.