Let
idSet denote the identity
functor on
Set and let
(−×Z) denote the
functor Set→Set given by
(−×Z):Set→Set:X↦X×Z:(f:X→Y)↦(f×idZ):X×Z→Y×Z
where
(g×h)(a,b)=(g(a),h(b)).
A natural transformation
idSet⇒(−×Z) consists of a bunch of component
morphisms. Let’s build two collections of component
morphisms and see what turns out to be natural.
We define our first as follows:
αX:X→X×ZαX(x)=(x,12)
and our second like this:
βX:X→X×ZβX={x↦(x,∣X∣)x↦(x,0)X is finiteotherwise
The key thing to notice is that the mapping
α acts “without regard to” what
X is, which is not true for
β. This will turn out to be what makes
α natural and not
β.
To find out for sure, we must check naturality. First
α; we need that
commute for all choices of
f. This holds pretty easily:
αb∘f=x↦(f(x),12)=x↦(f×id)(x,12)=(f×id)∘(x↦(x,12))=(f×id)∘αa
and hence, yes,
α is natural.
Now we check
β. Turns out that no,
β is not natural; we can construct a square that fails to commute. For
n∈N let
n={1,⋯,n}. Now consider the square
where
ι:4→8 is the
inclusion map. This square fails to commute because
(α8∘ι)(x)=(x,8)but
((ι×idZ)∘α4)(x)=(x,4)
Hence
β is not natural.
The point is that
β in some sense does something “different” on different choices of
X∈Set, and the above square for the
morphism ι:4→8 detects that it is
acting “differently” on
X=4 versus
X=8.
I have two somewhat incidental
notes to add to this example:
Any natural transformation
α:idSet⇒(−×Z) is determined by its component
α1, where
1 is the
terminal object of
Set. ... I think?.
Pretty sure natural transformations
α:idSet⇒(−×Z) are all of the form
α(x)=(x,n) for some fixed
n. This fact depends on
(−×Z) acting as
f↦f×idZ on
morphisms. What would happen with the alternative
functor (−×Z)′ which on
morphisms is, say,
f↦f×(−+1)?
Okay well that definition is not a
functor, because
(−×Z)(idA)=idA. But we could use this definition:
(−×Z)′′:FinSet→Set:A→A×Z:(f:A→B)→f×(z↦z+deg(f))
where
deg(f) is
defined as
deg(f:A→B)=−∣B∣+b∈B∑∣f−1({b})∣
The
function deg counts, in essence, how many values there are for which
f is not
injective.
Then
(−×Z)′′ is indeed a
functor, albeit from
FinSet instead of
Set.
Anyway, I wonder how this would interact with the example above? Do the natural transformations
idSet⇒(−×Z)′′ look meaningfully different from
idSet⇒(−×Z)
commute for all choices of . This holds pretty easily:
and hence, yes, is natural.
Now we check . Turns out that no, is not natural; we can construct a square that fails to commute. For let . Now consider the square
where is the inclusion map. This square fails to commute because
but
Hence is not natural.
The point is that in some sense does something “different” on different choices of , and the above square for the morphism detects that it is acting “differently” on versus .