This question deals with certain morphisms and 2-morphisms in the definition of coslice categories in 2-categories. Though I'm aware of the notion of lax (co)slice categories, I believe the following problem is distinct.

Let $\mathcal{C}$ be a 2-category and $X$ be an object of $\mathcal{C}$. The coslice category $X \downarrow \mathcal{C}$ has morphisms $f\colon X \to A$ for objects, and has for morphisms $N\colon (f\colon X \to A) \to (f'\colon X \to B)$ the morphisms $N\colon A \to B$ such that $f'=Nf$.

Assume now that we have $N_1\colon A \to B$, $N_2\colon A \to B$, with $f'=Nf$, $f''=Nf$, and that we also have a natural transformation $\alpha\colon N_1 \to N_2$. Intuitively, this would define some kind of morphism between the objects $f'\colon X \to B$ and $f''\colon X \to B$, but this is not the definition of 1-morphisms in $X \downarrow \mathcal{C}$.

The question is: what extra structure is brought to the category $X \downarrow \mathcal{C}$ by the natural transformations $\alpha\colon N_1 \to N_2$ ? I am thinking that $X \downarrow \mathcal{C}$ might be a double category, but can't quite define it precisely.

  • $\begingroup$ I'm not sure that this is entirely the right notion for the coslice 2-category. It depends on whether you want it to represent a weak of strong 2-limit. Try ncatlab.org/nlab/show/slice+2-category for a description of the dual notion in the weaker setting. $\endgroup$ – Tyrone Jul 12 '17 at 13:13
  • $\begingroup$ A natural candidate for a $2$-cell in $X \downarrow \mathcal{C}$ from $N_1 : A \to B$ to $N_2 : A \to B$ would be a $2$-cell $\alpha : N_1 \Rightarrow N_2$ in $\mathcal{C}$ such that $\alpha \cdot f = \mathrm{id}_{f'}$, where $\alpha \cdot f$ is an instance of whiskering. $$~$$ [P.S. since $2$-cells must be between parallel $1$-cells, your $f'$ and $f''$ should be equal, and $N_1,N_2$ are both $1$-cells in $X \downarrow \mathcal{C}$ from $f$ to $f'$.] $\endgroup$ – Clive Newstead Jul 12 '17 at 16:06

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.