Regularity Enforces Trivial $\in$-Isomorphisms

Question

Theorem 6.7. in Jech's Set Theory:

Assume Regularity. Let $T_1, T_2$ be transitive classes and let the bijection $\pi \colon T_1 \to T_2$ satisfy $u \in v \implies \pi u \in \pi v$. Then $\pi u = u$ for all $u \in T_1$ (and hence $T_1 = T_2$).

I have proved this statement independently and differently to Jech's proposed argument; of course I also want to understand his proof:

By $\in$-induction. Suppose $\pi z = z$ for all $z \in x \in T_1$. Then $x \subseteq \pi x$ because $z = \pi z \in \pi x$. Let $y \in \pi x$. Since $T_2$ is transitive, there is some z for which $\pi z = y$. Since $\pi z \in \pi x$, we have $z \in x$, and so $y = \pi z = z$. Thus $y \in x$.

I absolutely fail to see why the bold assertion follows (what obvious detail am I missing?). Could someone please point me in the right direction?

Answer 1

You have an error in the statement of the theorem. It should say $u \in v \iff \pi u \in \pi v$ instead of $u \in v \implies \pi u \in \pi v$. The line in question then follows immediately from this assumption.

The theorem is not true the way you wrote it. For instance, let $T_1=\{\emptyset,\{\emptyset\}, \{\{\emptyset\}\}\}$ and $T_2=\{\emptyset,\{\emptyset\}, \{\emptyset,\{\emptyset\}\}\}$ and let $\pi(\emptyset)=\emptyset$, $\pi(\{\emptyset\})=\{\emptyset\}$, and $\pi(\{\{\emptyset\}\})=\{\emptyset,\{\emptyset\}\}$.

Answer 2

By your choice of $y$ and $z$, you have $y\in\pi(x)$ and $\pi(z)=y$. Plug the latter into the former to conclude $\pi(z)\in\pi(x)$.

The $\implies$ in the assumption for $\pi$ must be a typo for $\iff$. Otherwise this would be a counterexample to the claim: $$ T_1 = \{\varnothing,\{\varnothing\},\{\{\varnothing\}\}\} \\ T_2 = \{ \varnothing,\{\varnothing\},\{\varnothing,\{\varnothing\}\}\} $$ with the "obvious" $\pi$.

Therefore $\pi(z)\in\pi(x)$ implies $z\in x$.