I use this notation: a well ordered set $Y$ is an ordinal if for every $a\in Y$, $Y_a=a$, where $Y_a=\{y\in Y|y< a\}$. Now, I know that for every woset there is an isomorphism from that woset to a (unique) ordinal. So let $f:X\longrightarrow Y$ be an iso, with $X$ a well-ordered set and $Y$ an ordinal. I claim that $X$ is an ordinal. Here is my proof: for every $a\in X$ we have
$$X_a=X_{f^{-1}(b)}=f^{-1}(Y_b)=f^{-1}(b)=a$$
Since I know that this is not the case, can you tell me where is the error in my proof?
The error lies in a confusion between taking the pre-image of $f$ on a point, versus taking the pre-image of a subset of the target. This is a subtle point that often arises with ordinals, since every ordinal $\beta$ below another ordinal $\alpha$ is both an element of $\alpha$ as well as a subset of $\alpha$, and the pre-image of $\beta$ under a function will not generally agree for these two meanings.
To explain precisely, in the beginning of your argument, I assume that you mean $f^{-1}(b)=a$ in the sense that $b=f(a)$, that is, in the sense of a single point. But in the equality $f^{-1}(Y_b)=f^{-1}(b)$, you are taking the pre-image using $b$ as a subset of $Y$. So this is a different meaning for $f^{-1}(b)$, and it will not be equal to $a$, but to $X_a$.