I'm probably just missing something obvious but suppose that $T \subset 2^{< \omega}$ is a perfect tree with no terminal nodes (what about just $[T]$ non-empty?). If $Y \leq_{T} X$ for all $X \in [T], X \leq_T T$ must we have $Y \leq_T Z$ for all $Z \in [T]$? In other words, is the lower bound of the $T$ computable paths equal to the lower bound of all paths?

More generally, does the following hold for all $T, n, Y$? $$ \forall X\left[X \in [T] \land X \leq_T T \implies Y \leq_T X^{(n)} \right] \implies \forall Z\left[Z \in [T] \implies Y \leq_T Z^{(n)} \right] $$

I suspect the answer is yes but, if not, would it hold if we assume that $Z$ must be generic relative to $T$ (that was what initially motivated the question and then I realized I didn't know the answer more generally)?

I'm probably just missing something obvious but suppose that $T \subset 2^{< \omega}$ is a perfect tree with no terminal nodes (what about just $[T]$ non-empty?). If $Y \leq_{T} X$ for all $X \in [T], X \leq_T T$ must we have $Y \leq_T Z$ for all $Z \in [T]$? In other words, is the lower bound of the $T$ computable paths equal to the lower bound of all paths?

More generally, does the following hold for all $T, n, Y$? $$ \forall X\left[X \in [T] \land X \leq_T T \implies Y \leq_T X^{(n)} \right] \implies \forall Z\left[Z \in [T] \implies Y \leq_T Z^{(n)} \right] $$

I suspect the answer is yes but, if not, would it hold if we assume that $Z$ must be generic relative to $T$ (that was what initially motivated the question and then I realized I didn't know the answer more generally)?

Edit: by generic relative to T I meant being generic in the local forcing on T (perhaps relative to the degree of T).

I'm probably just missing something obvious but suppose that $T \subset 2^{< \omega}$ is a perfect tree with no terminal nodes (what about just $[T]$ non-empty?). If $Y \leq_{T} X$ for all $X \in [T], X \leq_T T$ must we have $Y \leq_T Z$ for all $Z \in [T]$? In other words, can the paths which $T$

More generally, does the following hold for all $T, n, Y$? $$ \forall X\left[X \in [T] \land X \leq_T T \implies Y \leq_T X^{(n)} \right] \implies \forall Z\left[Z \in [T] \implies Y \leq_T Z^{(n)} \right] $$

I suspect the answer is yes but, if not, would it hold if we assume that $Z$ must be generic relative to $T$ (that was what initially motivated the question and then I realized I didn't know the answer more generally)?

I'm probably just missing something obvious but suppose that $T \subset 2^{< \omega}$ is a perfect tree with no terminal nodes (what about just $[T]$ non-empty?). If $Y \leq_{T} X$ for all $X \in [T], X \leq_T T$ must we have $Y \leq_T Z$ for all $Z \in [T]$? In other words, is the lower bound of the $T$ computable paths equal to the lower bound of all paths?

More generally, does the following hold for all $T, n, Y$? $$ \forall X\left[X \in [T] \land X \leq_T T \implies Y \leq_T X^{(n)} \right] \implies \forall Z\left[Z \in [T] \implies Y \leq_T Z^{(n)} \right] $$

I suspect the answer is yes but, if not, would it hold if we assume that $Z$ must be generic relative to $T$ (that was what initially motivated the question and then I realized I didn't know the answer more generally)?