Is there only one natural transformation from the product functor to itself? - MathOverflow

Is there only one natural transformation from the product functor to itself?

2012-08-27T13:23:29Z

Say we are working in a category which has all binary products. I guess that the identity transformation is the only natural transformation from $\times$ to $\times$. Is this really the case? If yes, how can I prove this? If not, does it help to assume that the category is cartesian closed and has all coproducts? What about natural transformations from $+$ to $+$ in CCCCs? Are they (also) unique?

Answer by Chris Schommer-Pries for Is there only one natural transformation from the product functor to itself?

2012-08-27T13:45:53Z

No, this is not the case.

Let the category C be vector spaces (say over the real numbers). Given any real number we get a natural transformation of the identity functor on C. On components for a given vector space V, this transformation is defined to be multiplication by the given real number.

If we whisker this (on the target side) with the product functor we get an infinite family of natural transformations from $\times$ to $\times$.

You can get a more exotic example by noting that a pair of real numbers gives an automorphism of the identity functor of $C \times C$, hence by wiskering (on the source side) we get another family. In components this is the transformation which on $V \oplus W$ scales V by the first number and W by the second.

Answer by Todd Trimble for Is there only one natural transformation from the product functor to itself?

2012-08-27T14:13:45Z

The key issue is how many natural transformations there are from the identity functor on $C$ to itself. Chris Schommer-Pries observed that there are many such transformations for $C = Vect$, one for each scalar.

A product functor $\prod: C \times C \to C$ is right adjoint to the diagonal functor $\Delta: C \to C \times C$, and it is easily seen that the collection of natural transformations $[\prod, \prod]$ is in natural bijection with the collection $[\Delta, \Delta]$, by a process called "taking the mate". Under the natural equivalence

$$(C \times C)^C \simeq C^C \times C^C$$

the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection

$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$

So in situations where there is at most one natural transformation from $1_C$ to itself, we get only one natural transformation from $\prod$ to itself.

Consider for example the case of cartesian closed categories $C$. If we compute the hom of $C$-enriched transformations, we can use the isomorphism $1_C \cong C(1, -)$ where $1$ on the right is the terminal object. By a Yoneda argument, the hom-object of enriched natural transformations is $C(1, 1) \cong 1$. (And the hom-set of enriched transformations would be $\hom_C(1, 1)$, which is a 1-element set.) More generally still, if $C$ is symmetric monoidal closed and $I$ is the monoidal unit, the set of enriched transformations from $1_C$ to itself will be in natural bijection with $\hom_C(I, I)$; this specializes to Chris's observation where we have $\hom_{Vect}(k, k) \cong k$.

Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection

$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$

and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself.

Edit: Of course, I still haven't given you an example of a cartesian closed category where the identity functor has more than one self-transformation (in the unenriched sense). But these are easy to come by. Consider for example the topos $Set^G$ of permutation representations of an abelian group $G$, i.e., functors $G \to Set$ where $G$ is regarded as a one-object category. Then for any $g \in G$, the action $g \cdot - : X \to X$ on $G$-sets $X$ is a $G$-set morphism that provides a natural transformation from the identity $Set^G \to Set^G$ to itself.