No, rigidity is not preserved in iterations. In particular,

**Proposition:** If $T$ is a rigid Souslin-tree with the property that
$$1 \Vdash_T T_{s} =\{ t \in T : t\le_T s \}\text{ is rigid and Souslin for every }s \not\in \dot{G}$$

(rigid and Souslin off-the-generic-branch in the terminology of FuchsHam2008.) Then,

- $1\Vdash_T $" $\check{T}$ is rigid and totally-proper."

**Proof:** Assume $T$ satisfies the required property.

To see that $T$ remains proper, fix some countable elementary sub-model $M\prec H(\lambda)$ with $T \in M$ and let $t\in T\cap M$ and $\dot{s} \in M^{T}$ be such that $t \Vdash \dot{s} \in \check{T}$. Noting that $T$ is c.c.c., we can find some $\dot{s}_0 \in M^T$ such that $t \Vdash_T \dot{s}_0 \le_T \dot{s}$ and $\dot{s}_0 \not\in \dot{G}$. Now let $\dot{D} \in M^{T}$ be a $T$-name for a dense-open subset of $T$ and $(u,\dot{v})\le (t,\dot{s}_0)$ be any extension with $Lev_{T}(u) \ge \delta = M\cap \omega_1$ and $u \Vdash Lev_{T}(\dot{v}) \ge \check{\delta}$.

Then, we must have $ u \Vdash (\exists r \in \dot{D} \cap \check{M})(\dot{v} \le r)$ (since otherwise, a standard reflection argument yields $u \Vdash (\exists a \in T \cap M)(a \not\in \dot{G}$ and $T_a$ is not Souslin$)$). It follows that $u \Vdash_T \dot{v}$ is totally $(M[\dot{G}], \check{T})$-generic; and so $1 \Vdash_{T} \check{T}$ is totally-proper.

To see that $T$ remains rigid, note that if $t \Vdash_{T} \check{T}$ is not rigid, then for some $s \in T$ with $s \perp t$, we must have $t \Vdash_{T} \check{T}_s$ is not rigid, or $\check{T}_s$ is not c.c.c. $\square$.

**Remark:** To see that this provides a counter-example, note that the two-step iteration of $T$ with itself is isomorphic to the square $T^2$ which admits the non-trivial automorphism $(s,t)\rightarrow (t,s)$. (being totally-proper in the extension didn't really matter, I just thought it was worth pointing out.)

think(but I'm not sure) it is still forcing-rigid in $L[a]$ where $a$ is $P$-generic over $L$. If so, then $P \times P$ is an example where the first p.o. is rigid in $L$, the second is rigid in the extension, but their composition (the product) is certainly not rigid, because if $(a,b)$ is a generic pair then so is $(b,a)$. It's all in chapter 28 of Jech if you want to take a look. $\endgroup$33more comments