1
$\begingroup$

Given a category $C$ with two objects and one non-identity morphism

$$a\to b$$

and another similar category $D$

$$x\to y,$$

we can define two functors $F,G:C\to D$ with

$$F:a\mapsto x, b\mapsto y$$

and

$$G:a\mapsto x, b\mapsto x$$

with morphisms doing the only thing they possibly can.

A natural transformation $\alpha:F\Rightarrow G$ would require a component $\alpha_b:F(b)\to G(b)$, but there is no morphism $y\to x$, so if I understand this correctly, there is no natural transformation from $F$ to $G$.

Is that correct? Is there a clear set of criteria required for there to exist a natural transformation?

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ You are right: there is no natural transformation from $F$ to $G$. I can't see what Bruce's comments about identity morphisms are about, but I agree with him that this is hardly an MO-level question. $\endgroup$
    – Robin Chapman
    Commented Aug 16, 2010 at 6:47
  • $\begingroup$ Not sure what "MO level" means, but if my question is below the level expected here, then it should be easy to provide an answer, no? Sorry for being a bit confused by your responses. Like Bruce, I always assumed there was SOME trivial natural transformation between functors, e.g. sending everything to the identity, but this example clearly demonstrates that is not the case. So if natural transformations only exist under certain conditions, what are those conditions? $\endgroup$
    – EricForgy
    Commented Aug 16, 2010 at 7:06
  • 1
    $\begingroup$ Eric, in general you cannot expect there to be natural transformations between two functors $C\to D$. A simpler example to yours is to take $C$ to be the category with one object and one morphism and $D$ any category. Then a functor from $C$ to $D$ amounts to choosing an object of $D$. and a natural transformation amounts to a morphism between the two objects. So if you have two objects $x$ and $y$ in $D$ with no morphism from $x$ to $y$ then there is no natural transformation between the functors $F$ and $G$ which take the object of $C$ to $x$ and $y$ respectively. $\endgroup$
    – Robin Chapman
    Commented Aug 16, 2010 at 8:53
  • $\begingroup$ Thanks Robin. That is helpful. Still, at least for finite categories, I'd think there should be some simple criteria telling us when a naturally transformation exists. We now even have a couple ways to cook them up at will. $\endgroup$
    – EricForgy
    Commented Aug 16, 2010 at 9:40
  • $\begingroup$ Sorry. I should wear my reading glasses. $\endgroup$
    – Bruce Westbury
    Commented Aug 16, 2010 at 19:14

1 Answer 1

Reset to default
3
$\begingroup$

In general, there are probably no conditions for the existence of natural transformations that are simpler than just using the definition of naturality itself. In a category Z with a zero object, the zero natural transformation between functors $F,G : C \rightarrow Z$ always exists, but this is quite a degenerate example.

Often, you can arrive at an intuition as to whether a natural transformation exists if you consider what such a thing would mean in the specific categories involved. For instance, functors from $(\bullet \rightrightarrows \bullet)$ into Set are just graphs. A natural transformation $G\Rightarrow H$ then exists precisely when there is a graph homomorphism from $G$ to $H$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Another useful case is when D is a groupoid of isomorphic objects --- you might call it a connected groupoid; the proof is related to that of how every category is equivalent to a skeletal subcategory. $\endgroup$
    – some guy on the street
    Commented Aug 16, 2010 at 16:20

You must log in to answer this question.

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