If $(X, \lt)$ is a well-ordering I can show by transfinite recursion over the ordinals that the function $f(x) = \text{ran} f |_{\hat{x}}$ exists (where $\hat{x} = \{ y : y \lt x\}$).
I have obtained $f$ this way, Let $V$ be the class of all sets and $F:V \to V$ be a class-function, then there is a unique $G:ON \to V$ where $ON$ is the class of all ordinals such that $F(\alpha) = F(G|_\alpha)$. So I can apply this to get a function $f$ such that $f(x) = F(f|_\hat{x})$. Now I let $F = \{(x, \text{ran} x) : x \in V\}$ and then I get the function as above.
Now, this should be an isomorphism (order preserving bijection) between $X$ and the set of true initial segments of $X$, $I_X$ ordered by inclusion.
However, when I have $x < y$, then I see that $\text{ran} f|_\hat{x} \subset \text{ran} f|_\hat{y}$. So $f(x) \leq f(y)$. Why do I have $f(x) \neq f(y)$?