Suppose we have two categories $D$ and $S$, as well as two functors $K,L:D\to S$ and a natural transformation $\varphi:K\to L$.

Given another category $C$ and a functor $Y:C\to S^D$, is there a nice way to denote the functor from $(Y\downarrow K)$ to $(Y\downarrow L)$ induced by $\varphi$?

I'll expose the motivation behind this question: Suppose that $K,L:\mathbf\Delta^{op}\to\text{Set}$ are simplicial sets, and $\varphi:K\to L$ is a morphism in the category of simplicial sets. I think we can define the simplex category of $K$ as the comma category $$\Delta\downarrow K$$ where $\Delta:\mathbf\Delta\rightarrow\text{sSet}$ is the Yoneda embedding for $\mathbf\Delta$ defined by $\phi\mapsto\hom_{\mathbf\Delta}(-,\phi)$ for an object or morphism $\phi$ of $\mathbf\Delta$.

I'd like to find an elegant way to denote the functor from $(\Delta\downarrow K)$ to $(\Delta\downarrow L)$ induced by $\varphi$. However, I'm not sure if the notation $(\Delta\downarrow\varphi):(\Delta\downarrow K)\to(\Delta\downarrow L)$ makes any formal sense!

Any ideas?

  • 2
    $\begingroup$ $(\Delta \downarrow \phi)$ is not unreasonable. $\endgroup$
    – Zhen Lin
    Sep 14, 2014 at 20:39
  • $\begingroup$ I like using $F(f)$ to denote the action of a functor $F$ on a morphism $f$ (not just on objects), and this seems consistent with that convention. $\endgroup$ Sep 14, 2014 at 21:24

1 Answer 1


In fact, your construction works for any small category $\mathscr A$ for which there is a functor $$ \widehat{\mathscr A} \to \mathsf{Cat},\, F \mapsto {\mathscr A}\big/{F} $$ where ${\mathscr A}\big/{F}$ is just another notation for the comma category $(\mathbf{yon}_{\mathscr A}\downarrow F)$. (This notation is justified by viewing $\mathscr A$ as a subcategory of $\widehat{\mathscr A}$ through Yoneda's lemma.)

Grothendieck denotes this functor $i_{\mathscr A}$ in Pursuing Stacks.

So in your example, $i_{\boldsymbol\Delta}(\varphi)$ would be historically consistent. (Although not very intuitive, I agree.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .