Leinster's proof of "Representables preserve limits"

Question

Leinster (p.148) gives the following proof of the fact that representatives preserve limits:

I understand the argument, but why does this prove the claim? To prove that limits are preserved, one has to prove that if $(p_i:B\to D(i))_{i\in I}$ is a limit cone on $D$ in $\mathscr A$, then $(\mathscr A(A,p_i):\mathscr A(A,B)\to \mathscr A (A,D(i)))_{i\in I}$ is a limit cone on $\mathscr A(A,D(-))$ in $\text{Set}$ (Definition 5.3.1). How exactly does it follow from the given sequence of isomorphisms?

Answer 1

Leinster's being slick here, which as you are experiencing is not very satisfying. This is actually an interesting result. To see what he's saying it might help to look at a direct proof which I'll outline.

Let $F: J \to \mathcal{C}$ be a diagram with a limiting object $\text{Lim } F$ equipped with the morphisms $\sigma_i: \text{Lim } F \to F_i$. Then applying the $\text{Hom}_{\mathcal{C}}(C, -)$ functor to $\text{Lim } F$ and to each $u_i$, we realize it forms a cone in $\textbf{Set}$. Now we show that $\text{Hom}_{\mathcal{C}}(C, \text{Lim } F)$, equipped with the morphisms $\sigma_{i*}$, is a universal cone; that is, it is a limit. Suppose that $X$ is a set which forms a cone with the morphisms $\tau_i: X \to \text{Hom}_{\mathcal{C}}(C, F_i)$. Then for each $x \in X$, we see that $\tau_i(x) : C \to F_i$. The diagram above tells us that $u \circ \tau_i(x) = \tau_j(x)$ for each $x$. Hence each $x \in X$ induces a cone with apex $C$ with morphisms $\tau_i(x): C \to F_i$. (This is like the first isomorphism Leinster uses, because note that you could go from the third diagram to the second, just as we went from the second to the third right now). However, $\text{Lim } F$ is the limit of $F: J \to \mathcal{C}$. Therefore, there exists a unique arrow $h_x: C \to \text{Lim } F$ such that $h_x \circ \sigma_i = \tau_i(x)$. Now we can uniquely define a function $h: X \to \text{Hom}_{\mathcal{C}}(C, \text{Lim } F)$ where $h(x) = h_x: C \to \text{Lim } F$, in such a way that the diagram below commutes.
So we see that the Hom functor does in fact preserve limits. The second isomorphism that Leinster is just him reinterpreting what a cone really means in this situation. But when Leinster wraps it up in that esoteric notation it's of course not clear, so diagrams help more with this stuff than notation.

Answer 2

Here's another direct proof. I'll use Leinster's notation and the following definition of limit preservation (the proof is essentially given here):

A functor $F: \mathscr A\to\mathscr B$ preserves limits if the following property is satisfied: whenever $D:I\to\mathscr A$ is a diagram that has a limit $(\lim D,p_i:\lim D\to D(i))_{i\in I}$, the composite $F\circ D: I\to\mathscr B$ also has a limit $(\lim(F\circ D), q_i: F\circ D\to FD(i))_{i\in I}$, and the unique arrow $\alpha: F(\lim D)\to\lim (F\circ D)$ with the property $p_i=q_i\circ \alpha$ for all $i\in I$ (whose existence is guaranteed by the definition of the limit of $F\circ D$) is an isomorphism.

Assume $D:I\to \text{Set}$ is a diagram that has a limit $(\lim D, p_i:\lim D\to D(i))_{i\in I}$.

By Example 5.1.22 in Leinster, all limits in $\text{Set}$ exist and are explicitly described in that example. In particular, the limit of the functor $\mathscr{A}(A,D(-)):I\to \text{Set}$ exists. The vertex of this limit cone, which we call $\lim \mathscr A(A,D(-))$, is the set $$\{(x_i)_{i\in I}: x_i\in \mathscr A(A,D(i))\text{ and for all }u:i\to j\text{ in } D,\ \mathscr A(A,D(u))(x_i)=x_j\}$$ and the projections are given by $q_i:(x_i)_{i\in I}\mapsto x_i$.

Now, according to the definition quoted above, it remains to show that the unique arrow $\alpha: \mathscr A(A,\lim D)\to \lim\mathscr A(A,D(-)) $ such that $q_i\circ\alpha=\mathscr A(A, p_i)$ for all $i\in I$ is an isomorphism.

Let's construct an inverse to $\alpha$, call it $\beta$. By Lemma 6.2.1 (or by inspection), $\lim \mathscr A(A,D(-))\simeq \text{Cone}(A,D)$, so we can regard the elements of $\lim \mathscr A(A,D(-))$ as cones on $D:I\to\mathscr A$ with vertex $A$. Define $\beta$ as follows. Assign to a cone $(A,x_i:A\to D(i))_{i\in I}$ the unique arrow $\Gamma: A\to \lim D$ such that $p_i\circ \Gamma=x_i$. Let's show that $\alpha\circ\beta=1,\ \beta\circ\alpha=1$.

First, $\alpha\circ\beta=1$. Note that $\beta\circ\alpha: (x_i)_{i\in I}\mapsto \Gamma\mapsto (p_i\circ \Gamma)_{i\in I}$. Since we must have $p_i\circ \Gamma=x_i$ by the definition of $\beta$, this implies $\alpha\circ\beta=1$.

Now for $\beta\circ \alpha=1$. Note that $\beta\circ\alpha: f\mapsto (p_i\circ f)_{i\in I}\mapsto \Delta$, where $\Delta$ is the unique arrow such that $p_i\circ \Delta=p_i\circ f$ for all $i\in I$. Finally, note that we must have $\Delta=f$ by uniqueness of the arrow, so $\beta\circ \alpha=1$.

Answer 3

Leinster's comment suffices due to the following:

Fix a diagram D, and suppose we have two cones $(A \xrightarrow{a_i} \mathcal{A}(A,Di))_{i \in I}, (L \xrightarrow{l_i} \mathcal{A}(A,Di))_{i \in I}$ on the same diagram. Suppose additionally that $(L \xrightarrow{l_i} \mathcal{A}(A,Di))_{i \in I}$ is a limit cone. I claim that if L and A are isomorphic, then $(A \xrightarrow{a_i} \mathcal{A}(A,Di))_{i \in I}$ is also a limit cone.

If this is true, it is easy to see why Leinster's proof suffices. We have that the vertices of the two cones are isomorphic by using the A-th component of the natural transformation given.

I will sketch the proof. Let $f:A \rightarrow L$ be our isomorphism and let $(B \xrightarrow{b_i} \mathcal{A}(A,Di))_{i \in I}$ be any cone on D. B factors uniquely through L via $b: B \rightarrow L$. I claim the composition $ b \circ f^{-1}$ is unique and satisfies the universal mapping property of limit cone. Uniquesness follows from uniqueness of the maps b,f and the uniqueness of inverse. UMP follows from the universal mapping properties of b and application of f. Hence proved.