Q:

Proof of Hartogs' Lemma

I'm reading "Sets, Models and Proofs" by I. Moerdijk and J. van Oosten to have some basic understanding of set theory.
There's one step in "Hartogs' Lemma" proof that I don't see how to it works: I don't understand why 
$\{\alpha \in P/\mathord{\cong} \mid \alpha < [L]\}$
is isomorphic to $L$?
For sure each $\alpha$ is mapped to an $l\in L$, $\alpha \rightarrow L$ is injective. But, how could one assert it is also a surjective?
Below is quoted from "Sets, Models and Proofs":
Lemma 1.4.1 (Hartogs’ Lemma) For every set $X$ there is a well-ordering
$(L_X,\leq)$ such that there is no injective function from $L_X$ to $X$.
Proof. Let $P$ be the set of all pairs $(L,\leq)$ where $L$ is a subset of $X$ and $\leq$
is a well-ordering on $L$. We shall denote the pair $(L,\leq)$ simply by $L$.
For two such $L$ and $M$, we write $L\preceq M$ if there is a (necessarily unique,
by Proposition 1.3.7) embedding of well-ordered sets: $L\rightarrow M$. Note:

we have both $L\preceq M$ and $M\preceq L$, if and only if $L \cong M$;
if $L \cong L′$ and $M\cong M′$, then $L\preceq M$ if and only if $L′\preceq M′$.

We can therefore define an order relation $\leq$ on the set $P/\mathord{\cong}$ of equivalence
classes of $P$ modulo the equivalence relation $\cong$. By Proposition 1.3.9, the
set $P/\mathord{\cong}$ is a linear order with the relation $\leq$.
Note, that if $L \prec M$ (that is, $L\preceq M$ but $M \ncong L$), there is a unique
$m \in M$ such that $L$ is isomorphic to the set $\downarrow(m) = \{m′ \in M \mid m′ < m\}$
with the restricted order from $M$.
Therefore, if we denote the $\cong$-equivalence class of $L$ by $[L]$, the set
$$\{\alpha \in P/\mathord{\cong} \mid \alpha < [L]\}$$
is isomorphic to $L$.
Now suppose that $W \subseteq P/\mathord{\cong}$ is a nonempty set of $\cong$-equivalence classes.
Let $\alpha = [L]$ be an arbitrary element of $W$. Consider the set
$$L_W = \{l \in L\mid [\downarrow(l)] \in W\}$$
If $L_W$ is empty, clearly $[L]$ is the least element of $W$. If $L_W$ is nonempty,
then it has (as subset of the well-ordered set $L$) a least element $l_W$. But
then $[\downarrow(l_W)]$ is the least element of $W$. So every nonempty subset of $P/ \mathord{\cong}$
has a least element, and therefore $P/\mathord{\cong}$ is a well-ordered set.
There cannot be an injective function from $P/\mathord{\cong}$ into $X$, for suppose $f$
is such a function. Then $f$ gives a bijective function between $P/\mathord{\cong}$ and a
subset $Y_f$ of $X$; we can then give $Y_f$ the same well-ordering as $P/\mathord{\cong}$, so we
have $(Y_f ,\leq) \cong (P/\mathord{\cong},\leq)$. This is impossible however, since $[(Y_f ,\leq)]$ is an
element of $P/\mathord{\cong}$ (see Proposition 1.3.7).
--- end of quotation

A:

If you think about the von Neumann ordinal assignment then this becomes much more apparent:
An ordinal $\alpha$ is the set of all ordinals $\beta$ such that $\beta<\alpha$.

A:

As you noted, the previous paragraph

...if $L \prec M$ (that is, $L\preceq M$ but $M \ncong L$), there is a unique
  $m \in M$ such that $L$ is isomorphic to the set $\mathord{\downarrow}(m) = \{m′ \in M \mid m′ < m\}$
  with the restricted order from $M$.

implies that there is a canonical injection (call it $f$) mapping into $L$ from the set of all $\cong$-classes less than $[L]$. 
Namely, to a lesser $\cong$-class $[L'] < [L]$, the function
$f$ associates the unique element $l \in L$ such that the corresponding initial segment $\mathord{\downarrow} (l)$ of $L$ is isomorphic to $L'$.  In other words, $\mathord{\downarrow} f([L'])  \cong L'$ .
Once we have been convinced of this, the surjectivity of $f$ follows easily:  Let $l \in L$. Then
$f(\mathord{\downarrow} (l))$ is defined as the unique element $l'$ of $L$ such that 
$\mathord{\downarrow} (l') \cong \mathord{\downarrow} (l)$.  But clearly this unique element $l'$ is $l$ itself; that is, $f(\mathord{\downarrow} (l)) = l$.
By the way, I think it might be easier (and less confusing) to instead consider a map going the other way, taking every element $l \in L$ to the $\cong$-class of the initial segment $\mathord{\downarrow} (l)$ of $L$ that it determines, and to show that this map is a bijection.

