Talk:Condition for Woset to be Isomorphic to Ordinal

Hello, I think I found a mistake on this page but I'm not sure (I'm a beginner).

After the words "we have" one sees
 * $(2): \quad \map {g_y} {S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$ is an order isomorphism

I assume that the order isomorphism $\map{g_y}{S_x}$ should read $g_y\restriction S_x$. $\map{g_y}{S_x}$ stands for the image of $S_x$, a subset of $\map Z y$, which is not even a function (so it can't be a isomorphism). I believe it means a new, "smaller" isomorphism that comes down from $g_y$ - equal to $g_x$ - so a restriction should be the proper notation.

Is it indeed a mistake, or didn't I understand it right?


 * Seems to have been a mistake.


 * Should have read:


 * $(2): \quad \paren {g_y S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$ is an order isomorphism
 * where the notation probably means subset product. --prime mover (talk) 13:51, 7 August 2021 (UTC)


 * Thanks for the reply, but what has the topic here to do with binary operations? - Lutalli (talk) 14:52, 7 August 2021 (UTC)


 * I don't know. What do you understand the notation $\paren {g_y S_x}$ to be? --prime mover (talk) 15:25, 7 August 2021 (UTC)


 * If my understanding is right, this proof first gets the result (using the doubted notation)
 * $\paren {g_y S_x}: (S_y)_x \to \paren {\map Z y}_{\map {g_y} x}$
 * from
 * $g_y : S_y \to Z(y)$
 * and uses $S_x = (S_y)_x$ to obtain
 * $\paren {g_y S_x}: S_x \to \paren {\map Z y}_{\map {g_y} x}$
 * $(S_y)_x$ is an initial segment of $S_y$ given by $x$; $\paren {\map Z y}_{\map {g_y} x}$ is an initial segment of ${\map Z y}$ given by $g_y(x)$; and what the symbol $\paren {g_y S_x}$ wanted to express should be a corresponding "cut-down" of $g_y$, which can naturally be $g_y\restriction S_x$ or simply $g_x$ (the domain of the isomorphism is reduced from $S_y$ to $S_x$), as I said. -Lutalli (talk) 16:19, 7 August 2021 (UTC)