Is there only one natural transformation from the product functor to itself? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T15:24:08Z http://mathoverflow.net/feeds/question/105624 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105624/is-there-only-one-natural-transformation-from-the-product-functor-to-itself Is there only one natural transformation from the product functor to itself? Wolfgang Jeltsch 2012-08-27T13:23:29Z 2012-08-27T17:13:38Z <p>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?</p> http://mathoverflow.net/questions/105624/is-there-only-one-natural-transformation-from-the-product-functor-to-itself/105628#105628 Answer by Chris Schommer-Pries for Is there only one natural transformation from the product functor to itself? Chris Schommer-Pries 2012-08-27T13:45:53Z 2012-08-27T13:52:39Z <p>No, this is not the case. </p> <p>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.</p> <p>If we whisker this (on the target side) with the product functor we get an infinite family of natural transformations from $\times$ to $\times$. </p> <p>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. </p> http://mathoverflow.net/questions/105624/is-there-only-one-natural-transformation-from-the-product-functor-to-itself/105630#105630 Answer by Todd Trimble for Is there only one natural transformation from the product functor to itself? Todd Trimble 2012-08-27T14:13:45Z 2012-08-27T17:13:38Z <p>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. </p> <p>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 <a href="http://ncatlab.org/nlab/show/mate" rel="nofollow">mate</a>". Under the natural equivalence </p> <p>$$(C \times C)^C \simeq C^C \times C^C$$ </p> <p>the functor $\Delta$ is taken to the pair of identity functors, and we get under this equivalence a bijection </p> <p>$$[\Delta, \Delta] \cong [1_C, 1_C] \times [1_C, 1_C].$$ </p> <p>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. </p> <p>Consider for example the case of cartesian closed categories $C$. If we compute the hom of <em>$C$-enriched</em> 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-<em>set</em> 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 $\hom_{Vect}(k, k) \cong k$. </p> <p>Or, if $C$ has a terminal object $1$ and $\hom(1, -): C \to Set$ is faithful, we have an injection </p> <p>$$[1_C, 1_C] \to [\hom(1, -), \hom(1, -)]$$ </p> <p>and then an ordinary $Set$-based Yoneda argument shows there is at most one transformation from $1_C$ to itself. </p> <p><b>Edit:</b> 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 <em>unenriched</em> 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. </p>