Modified version of generelized Hechler forcing I'm looking at a modified version of the generalized Hechler forcing:
Let $\kappa$ be inaccessible, and fix some $s:\kappa \to \kappa$ strictly monotone taking value in the regular cardinals.* Conditions in the forcing are of the form $(f,g)$ where for some $\alpha<\kappa$, $f$ is a function from $\alpha$ to $\kappa$, $g$ is a function from $\kappa \smallsetminus \alpha$ to $\kappa$, and for all $\xi < \kappa$, either $s(\xi) > f(\xi)$ or $s(\xi) > g(\xi)$ (whichever of them makes sense). A condition $(f,g)$ is stronger then $(f',g')$ if $f' \subset f$ and for all $\xi \in \mathrm{dom}{g'}$ either $f(\xi) \ge g'(\xi)$ or $g(\xi) \ge g'(\xi)$ (again, whichever of them makes sense).
If we remove the restriction given by $s$, the forcing adds a function dominating all functions from $V$ in $^\kappa\kappa$ (mod bounded).
Say we force with the restriction given by $s$. Then the forcing adds a function dominating all functions from $V$ in $\prod_{i<\kappa}s(i)$. Does it also add a dominating function for $^\kappa\kappa$? What about the other way around - when we force without restricting using $s$, do we also add a dominating function for $\prod_{i<\kappa}s(i)$?
* It might be similar if we take $\kappa$ to be any regular cardinal, and ask that $s$ takes values in limit ordinals. But I defined it this way because I'm not sure, and because the broader setting  in which this question came up has $\kappa$ being supercompact.
 A: The restricted version never adds a dominating function in $\kappa^\kappa$ (mod bounded) (Edit: if $\kappa$ is weakly compact) and it is consistent (relative to a Mahlo cardinal) that the unrestricted forcing does not add a dominating function in $\prod_{i<\kappa}s(i)$ (mod bounded).
Lets first check the first part. Let us call the restricted by $s$ forcing $\mathbb P_s$. The crucial points are that

*

*Any $(f, g), (f^\prime, g^\prime)\in\mathbb P_s$ with $f=f^\prime$ are compatible.

*For any $\alpha<\kappa$ there are ${<}\kappa$-many $f$ with $\mathrm{dom}(f)\subseteq\alpha$ such that $(f, g)\in\mathbb P_s$ for some $g$.

Note that the second point is not true for the unrestricted forcing. Now let $\dot h$ be a name for a function $h:\kappa\rightarrow\kappa$. For $\alpha<\kappa$ let $\mathbb P_s^\alpha$ be the set of $(f, g)\in\mathbb P_s$ with $\mathrm{dom}(f)\subseteq\alpha$. The above implies that for any $i,\alpha<\kappa$ there are ${<}\kappa$-many $\beta$ such that some $(f, g)\in\mathbb P_s^\alpha$ decides $\dot h(\check i)$ to be $\beta$. Let us define
$$F_i:\mathbb P_s\rightarrow\kappa,\ F_i((f, g))=\mathrm{min}\{\alpha\mid \exists (f^\prime, g^\prime)\in\mathbb P_s^\alpha\text{ deciding }\dot h(\check i)\wedge (f^\prime, g^\prime)\leq (f, g)\}$$
and
$$H:\kappa\rightarrow\kappa,\ H(i)=\mathrm{sup}\{\beta\mid \exists (f, g)\in\mathbb P_s^{\mathrm{sup}F_i[\mathbb P_s^i]}\text{ deciding }\dot h(\check i)\text{ as }\beta\}+1$$
We must check that $H$ is well-defined. As j3M points out, it seems non-trivial that $F_i[\mathbb P_s^i]$ is always bounded in $\kappa$ (see the comments). We make use of the weak compactness of $\kappa$ to prove this:
So suppose $\mathrm{sup}F_j[\mathbb P_s^j]=\kappa$. There must then be a single first component $f$ so that $F_j((f, g))$ gets arbitrarily large by varying $g$. For convenience, say $f=\emptyset$. Let
$$X_\alpha=\{g\in\prod_{i<\kappa}s(i)\mid F_j(f, g)\geq\alpha\}$$
Observe that membership of $g$ in$ X_\alpha$ depends only on $g\upharpoonright\alpha$, so (abusing notation) we will say $g\upharpoonright\alpha\in X_\alpha$.
Consider the tree $T$ whose $\alpha$-th level $T_\alpha$ is exactly
$$T_\alpha=X_\alpha\cap\prod_{i<\alpha}s(i)$$
and is ordered by end-extension. Then $T$ is indeed a tree, all levels have size ${<}\kappa$ and by our choice of $f$, $T$ has height $\kappa$. By weak compactness, there is a cofinal branch through $T$. We may think of that branch as a function $g\in\bigcap_{\alpha<\kappa} X_\alpha$. But now $F_j((f, g))\geq\kappa$, meaning $(f, g)$ has no extension deciding the value of $\dot h(\check j)$. This is clearly a contradiction.
Finally $H$ is not dominated (mod bounded) by $\dot h^G$ for any generic $G$ (as the generic will often "pick minimal extensions deciding some value of $\dot h$").
For the second part, suppose that $\kappa$ is a Mahlo cardinal and $s(i)$ is simply the $i$-th regular cardinal. The set $S=\{i<\kappa\mid i=s(i)\}$ is stationary in $\kappa$ and note that any $f\in\prod_{i<\kappa}s(i)$ is regressive on $S$. Assume that after forcing with the unrestricted generalised Hechler forcing $\mathbb P$ at $\kappa$, there is $f\in\prod_{i<\kappa}s(i)$ dominating all such functions in $V$ (mod bounded). Then $f^{-1}(\{\xi\})$ is bounded and thus non-stationary for any $\xi<\kappa$. By Fodor's Lemma this means that $S$ is no longer stationary. However, $\mathbb P$ is ${<}\kappa$-closed and so preserves the stationarity of $S$, contradiction.
