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
As you noted, the previous paragraph
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.