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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?

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  • $\begingroup$ 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. $\endgroup$ – k.stm Jul 22 '19 at 7:35
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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.

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  • $\begingroup$ In the last paragraph (when you are saying "by the bijectivity of the map"), are you using surjectivity at all? I can only see that injectivity is used. $\endgroup$ – user634426 Jul 22 '19 at 21:14
  • $\begingroup$ 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. $\endgroup$ – jeanmfischer Jul 22 '19 at 21:21
  • $\begingroup$ I changed the word I used to be more precise. $\endgroup$ – jeanmfischer Jul 22 '19 at 21:25

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