1
$\begingroup$

Consider an arrow between the categories of endofunctors over two symmetric monoidal (SM) categories $\mathcal{C}$ and $\mathcal{D}$

$$a:End(\mathcal{C}) \rightarrow End(\mathcal{D})$$

It is a functor acting over endofunctors over a SM category but also acts over natural transformations between endofunctors over a SM category.

Is $a$ a 2-functor? Which (higher order) category $a$ belongs to (seen as an arrow)? Does $a$ send objects of $\mathcal{C}$ to objects of $\mathcal{D}$ in some sense?

$\endgroup$
  • $\begingroup$ There seems to be some confusion here. An arrow between two categories is usually a functor. A natural transformation is an arrow between functors. $\endgroup$ – Zhen Lin Apr 17 '14 at 9:28
  • $\begingroup$ Edited, thanks. $\endgroup$ – gibarian Apr 17 '14 at 9:44
1
$\begingroup$

Any object $C \in \mathcal{C}$ can be seen as a constant functor sending everything to the terminal category containing $C$, so yes, the objects of $\mathcal{C}$ are being sent to the objects of $\mathcal{D}$. And any morphism in $\mathcal{C}$ is a natural transformation of constant functors (easy exercise). So your $a$ gives rise to a functor $a: \mathcal{C} \to \mathcal{D}$.

But there is more to it. Consider the bicategory of symmetric monoidal categories. It has two 1-object subcategories, with objects $\mathcal{C}$ and $\mathcal{D}$, respectively. Your $a$ is a bifunctor between them. This is what you get if you "deloop" your picture of the categories $\operatorname{End}\mathcal{C}$ and $\operatorname{End}\mathcal{D}$, they're actually 1-object bicategories.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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