# Arrows $A\to L$ correspond to cones on $D$ with vertex $A$

I'm reading Remark 5.1.20(a) from Leinster (p. 119). He claims the following (in my words):

Fix a small category $$\mathbf I$$, a category $$\mathscr A$$, and a functor $$D:\mathbf I\to\mathscr A$$. Let $$(p_I:L\to D(I))_{I\in\mathbf I}$$ be a limit cone of $$D$$. Then for any $$A\in\mathscr A$$, there is a bijection $$\{\text{arrows } A\to L \text{ in }\mathscr A\}\leftrightarrow\{\text{cones on } D \text{ with vertex } A\}$$ given by assigning to $$g:A\to L$$ the cone $$(p_I\circ g:A\to D(I))_{I\in\mathbf I}$$. Leinster claims that the bijectivity of this assignment follows from the definition of limit. I'm trying to verify this in detail.

Surjectivity. I believe the statement is based on the assumption that a limit cone of $$D$$ exists. In such a case, any cone on $$D$$ with vertex $$A$$ gives rise to an arrow $$A\to L$$ (such that some properties hold, but this is not relevant here). So the assignment is surjective. Am I right here? Added after I've written the below passage on injectivity: actually this is not clear as well. We get an arrow from $$A$$ to a some limit $$L'$$ which need not a priori be equal to $$L$$...

Injectivity. This is where I don't understand what's going on. First of all, Leinster says that for now we do not know that a limit is unique and that it will follow from the claim being proved. So at this point we cannot use the uniqueness of a limit, right? To prove injectivity, we must prove that if $$(f_I:A\to D(I))$$ and $$(h_I:A\to D(I))$$ are two cones on $$D$$ with vertex $$A$$, then the arrows $$\overline f$$ and $$\overline h$$ from the definition of limit (5.1.19(b)) are equal. But a priori they are not even arrows to the same $$L$$ if we don't know that $$L$$ is unique. And the arrows $$p_I$$ from the definition of limit cone of $$(f_I)$$ need not be equal to the arrows $$p_I'$$ from the definition of limit cone of $$(h_I)$$. And even if we know that $$L$$ and $$p_I$$ are unique (which we don't), I still wasn't able to prove $$\overline h=\overline f$$. How to do that?

• You only need the uniqueness of a limiting arrow $A → L$ of cone vertices, making a given cone $C$ with vertex $A$ commute with the limit cone. – k.stm Jul 22 '19 at 7:35

You have to be careful : "L is a limiting cone over $$D$$" is a statement about $$L$$ ! You have supposed that $$L$$ is a limiting cone and so you have to use the properties of $$L$$ that follow from the definition of a limiting cone. You don't need to care about the possibility of other limiting cones for the moment.
We have that $$\{d_i : L \to D(i)\}$$ is a limiting cone over $$D$$ if for any cone $$\{h_i : C \to D(i)\}$$ there is a unique morphism $$h : C \to L$$ such that $$d_i \circ h = h_i$$. So here in the definition we have existence and unicity which should mean that there is a hidden bijection. What is the map realising the bijection ? Well it is the map that you described but I will write it in this way : $$Hom_{\mathcal A}(C,L) \to Nat_{Fun(I,A)}(\delta(C),D)$$ $$h \mapsto \{d_i \circ h, i \in I\}$$ Giving a cone with vertex $$C$$ is the same thing as giving a natural transformation between the constant functor $$\delta(C) : i\mapsto C$$ (which sends maps $$i \to j$$ in $$I$$ to the identity of $$C$$) and the diagram $$D$$.
The existence part of the definition gives surjectivity : given a cone $$\{h_i : C \to D(i)\}$$ there is a preimage $$h: C \to L$$ of the map just described, this is surjectivity. The unicity part is the injectivity of this map : Given two maps $$h : C \to L$$ and $$f : C\to L$$ such that $$d_i \circ f = d_i \circ g$$ for any $$i \in I$$, i.e. given two elements of $$Hom_{\mathcal A}(C,L)$$ which induce the same cone over $$D$$, then $$h = g$$ by unicity.
Then the unicity of $$L$$ follows : given $$\{ d_i' : L'\to D(i)\}$$ another limiting cone by definition there are two maps $$f :L \to L'$$ and $$h : L' \to L$$ such that $$d_i \circ h = d_i'$$ and $$d_i' \circ f = d_i$$, but then, $$h\circ f : L \to L$$ in $$Hom_{\mathcal A}(L,L)$$ will be sent to $$\{d_i \circ h \circ f = d_i, i \in I\}$$ meaning $$h\circ f = id_L$$ by the injectivity of the map. Similarly $$f\circ h = id_{L'}$$. Meaning that $$L$$ and $$L'$$ are isomorphic in $$\mathcal A$$, and they are actually isomorphic in a unique way as cones over $$D$$ meaning $$f$$ and $$h$$ are the unique maps that make the diagrams comute.
• I’m using surjectivity in defining the maps $f$ and $h$, then I’m using injectivity to compare the compositions to the identity which have the same image under the map. – jeanmfischer Jul 22 '19 at 21:21