Equivalent definitions of preserving limits On page 137, Leinster gives two equivalent characterizations of limit preservation:

Is it supposed to be obvious that they are the equivalent? If so, how to see that? (When I tried to prove that, I got stuck at the points which I ask about below.)
Also, I have two questions about his definition of 'canonical map'. First, it involves some $I$-components, which alludes to the fact that this is some kind of family of arrows. But how can it be a family if it is a specific arrow from one object of $\mathscr B$ to another? Second, when he defines what the $I$-component is, the target object of this $I$-component arrow is $F(D(I))$. Is it the same as $\lim(FD)$? If so, why?
 A: Generalities about limits, words and their definitions:
Your confusion about the canonical map is completely understandable. People are sometimes vague about what they mean by these things, since to an expert it's clear, but it can be difficult for the learner.
The key here is to remember the universal property of the limit. 
First to set things up, let $D:I\to \mathscr{A}$ be a diagram, $(X,\{f_i\}_{i\in I})$ be a cone to $D$, meaning that $X$ is an object of $\mathscr{A}$, which I will call the apex of the cone, each
$f_i :X\to Di$ is a morphism in $\mathscr{A}$, called the $i$th component of the cone, and $i\in I$ meaning that $i$ is an object of $I$, and for each morphism $u:i\to j$ in $I$, we have $f_j \circ Du = f_i$.
Then the universal property says that for each cone $\newcommand\set[1]{\left\{{#1}\right\}}(X,\set{f_i})$, there is a unique map $(X,\set{f_i})\to \lim D$.
 Note that I've written the map from the cone $(X,\set{f_i})$, this is because it's important to remember that the limit is itself a cone to $D$ $(Y,\set{g_i})$. We usually omit the morphisms $\set{g_i}$ when describing the limit, but they're important to remember for two reasons.


*

*The map $(X,\set{f_i})\to (Y,\set{g_i})$ in the universal property is a map $\alpha : X\to Y$ such that $g_i\circ \alpha = f_i$ for all $i\in I$.

*For any map $\alpha : X\to Y$, if we define $f_i = g_i\circ \alpha$, then $(X,\set{f_i})$ gives a cone to $D$ with universal map $\alpha: (X,\set{f_i})\to (Y,\set{g_i})$.


Thus we get a bijection between cones to $D$ with apex $X$ and maps from $X$ to $Y$, where $Y$ is the apex of the limit cone. Now remember that I defined the morphisms $\set{f_i}$ to be the components of the cone. Since we have this bijection, we say that the $i$th component of a map $\alpha:X\to Y$ is the $i$th component of the corresponding cone, which is $g_i\circ \alpha$. Note that the $i$th component is a map $X\to Di$.
The specific case
Let $(Y,\set{g_i})$ be the (actually a) limit cone to $D$ in $\mathscr{A}$.
Applying the functor $F$ gives us a cone $(FY,\set{Fg_i})$ to $F\circ D$ in $\mathscr{B}$.
Thus if $\lim F\circ D$ exists in $\mathscr{B}$, say it is $(Z,\set{h_i})$, with $Z\in\mathscr{B}$, $h_i : Z\to FDi$, the universal property of the limit gives a map 
$$\alpha : (FY,\set{Fg_i})\to (Z,\set{h_i}).$$
In other words a map 
$\alpha :FY \to Z$ such that $h_i\circ\alpha = Fg_i$. But remember that we defined the $i$th component of $\alpha$ to be exactly $h_i\circ\alpha$. So we could rephrase this as 
$\alpha$ is a map from $F(\lim D)\to \lim F\circ D$ such that the $i$th component is $Fg_i$.
This map is always defined if the limits exist regardless of whether or not $F$ preserves limits. Now we want to show that $F$ preserves this limit (meaning $(FY,\set{Fg_i})$ is a limit cone for $F\circ D$) if and only if this canonical map $\alpha$ is an isomorphism.
The equivalence

$(FY,\set{Fg_i})$ is a limit cone for $F\circ D$ if and only if the canonical map $$\alpha: F\lim D \to \lim F\circ D$$ 
  is an isomorphism.

Proof:
First suppose this map is an isomorphism. Let $(X,\set{f_i})$ be any cone to $F\circ D$.
We want to show that there is a unique map $\gamma:(X,\set{f_i})\to (FY,\set{Fg_i})$ such that $Fg_i \circ \gamma = f_i$. For existence, let $\beta : (X,\set{f_i})\to (Z,\set{h_i})$ be the canonical map from the universal property of $(Z,\set{h_i})$, such that $h_i \circ \beta = f_i$. Define 
$$ \gamma : X\xrightarrow{\beta} Z\xrightarrow{\alpha^{-1}} FY .$$
Then 
$$Fg_i \circ \gamma = Fg_i \alpha^{-1} \beta = h_i\alpha\alpha^{-1}\beta = h_i\beta = f_i,$$
as desired.
Conversely, suppose $\widetilde{\gamma}$ also had the property that 
$Fg_i \circ \widetilde{\gamma} = f_i$, then $\alpha\circ\widetilde{\gamma} : X\to Z$ has the same components as $\beta$, and therefore must equal $\beta$ by the uniqueness of $\beta$. Thus $\widetilde{\gamma}=\gamma$, so $\gamma$ is unique, and $(FY,\set{Fg_i})$ is a limit cone.
Now suppose $(FY,\set{Fg_i})$ is a limit cone. Then the universal property of limits says that there is a unique map $\alpha : (FY,\set{Fg_i})\to (Z,\set{h_i})$ (the canonical map), and there is also a unique map $\beta : (Z,\set{h_i})\to (FY,\set{Fg_i})$, and the composite $\beta\circ \alpha$ must be the identity of $(FY,\set{Fg_i})$ and 
$\alpha \circ \beta$ must be the identity of $(Z,\set{h_i})$ from the universal property of the limit. Hence $\alpha$ and $\beta$ are inverse isomorphisms. In particular $\alpha$ is an isomorphism, as desired. $\blacksquare$
