Assume $\kappa > \aleph\_1$ is regular and let $P=(P\_\alpha, \dot{Q}\_\beta: \alpha \leq \kappa^+, \beta< \kappa^+)$ be an iteration such that:


1) For even $\beta, \dot{Q}\_\beta$ is forced to be $\aleph\_1$-closed, $\kappa$-c.c. and a subset of $V\_\kappa$.


2) For odd $\beta, \dot{Q}\_\beta$ is forced to be $\kappa$-closed,


3) For $p \in P,$ if $Supp(p)$ denotes the support of $p$, then $|Supp(p) \cap E| < \aleph\_1$ and $|Supp(p) \cap O|< \kappa$ (where $E$ and $O$ are the class of all even and odd ordinals respectively).


From $P$ we can naturally define the forcing notions $P^E$ and $P^O$, where $$P^E=\{p \in P: \forall \beta \in O, p\restriction \beta \Vdash p(\beta)=1 \}$$
and
 $$P^O=\{p \in P: \forall \beta \in E, p\restriction \beta \Vdash p(\beta)=1 \}$$.
Note that there is a natural map $$\pi: P^E \times P^O \to P$$ which is defined as follows: $\pi(p, q)=r$, where for even $\beta, r(\beta)$
is forced to be $p(\beta)$ and for odd $\beta$ it is forced to be $q(\beta)$.


Clearly $P^O$ is $\kappa$-closed.



> 
> **Question.** Is $\pi$ a projection of forcing notions, assuming $P^E$ is $\kappa$-c.c.?
> 
> 
> 


**Remark.** I think the extra assumption ``$P^E$ is $\kappa$-c.c.'' is necessary for the question, as otherwise, maybe the arguments from "[The $\aleph\_2$-Souslin Hypothesis](http://www.jstor.org/stable/1998547?origin=crossref&seq=1#page_scan_tab_contents)" may be used to show that for some suitable choice of the forcings, $P$ may collapse $\kappa,$ while this can not happen for the product $P^E \times P^O$ by Easton's lemma.



Iterated forcing and projections
You can drop the hypothesis of $P^E$ being $\kappa$-c.c.


So take $(p\_1,p\_2)\in P^E\times P^O$, and set $p=\pi(p\_1,p\_2)$. Let $q\in P$ be such that $q\leq p$.


We shall build $q\_1\in P^E$ and $q\_2\in P^O$ such that $(q\_1,q\_2)\leq(p\_1,p\_2)$ and $\pi(q\_1,q\_2)\leq q$. We define $q\_1\upharpoonright\beta$ and $q\_2\upharpoonright\beta$ by induction on $1\leq\beta\leq\kappa^+$.


For $\beta=1$ set $q\_1\upharpoonright\beta=q\upharpoonright\beta$ and $q\_2\upharpoonright\beta=\{(0,1)\}$.


Suppose for some $\beta\leq\kappa^+$ both $q\_1\upharpoonright\alpha$ and $q\_2\upharpoonright\alpha$ have been defined for all $\alpha<\beta$, with $q\upharpoonright\alpha\leq q\_1\upharpoonright\alpha, q\_2\upharpoonright\alpha$ and $q\_i\upharpoonright\alpha\leq p\_i\upharpoonright\alpha$ for $i=1,2$.


If $\beta$ is odd, say $\beta=\gamma+1$, let $q\_1(\gamma)$ be a $ P\upharpoonright\gamma$-name such that $q\upharpoonright\gamma\Vdash q\_1(\gamma)=q(\gamma)$ and for all $q'\leq q\_1\upharpoonright\gamma$ in $ P\upharpoonright\gamma$ with $q'\perp q\upharpoonright\gamma$ we have $q'\Vdash q\_1(\gamma)=p(\gamma)$. Finally, let $q\_2(\gamma)$ be a $ P\upharpoonright\gamma$-name such that $q\_2\upharpoonright\gamma\Vdash q\_2(\gamma)=1$. It is easy to verify that $q\upharpoonright\beta\leq q\_1\upharpoonright\beta,q\_2\upharpoonright\beta$ and $q\_i\upharpoonright\beta\leq p\_i\upharpoonright\beta$.


Similarly, if $\beta$ is even and successor, we define $q\_2(\gamma)$ in such a way that $q\upharpoonright\gamma\Vdash q\_2(\gamma)=q(\gamma)$ and $q\_2\upharpoonright\gamma-q\upharpoonright\gamma\Vdash q\_2(\gamma)=p(\gamma)$, and $q\_1(\gamma)$ with $q\_1\upharpoonright\gamma\Vdash q\_1(\gamma)=1$, again we have $q\upharpoonright\beta\leq q\_1\upharpoonright\beta,q\_2\upharpoonright\beta$ and $q\_i\upharpoonright\beta\leq p\_i\upharpoonright\beta$.


For $\beta$ limit, $q\_i\upharpoonright\beta$ is just the union of $q\_i\upharpoonright\alpha$, $\alpha<\beta$.


We have $q\_i\in P$ as $supp(q\_i)\subseteq supp(q) \cup supp(p)$. 


It is clear that $(q\_1,q\_2)\leq (p\_1,p\_2)$.


It is easy to show by induction on $1\leq\beta\leq\kappa^+$ that $\pi(q\_1,q\_2)\upharpoonright\beta\leq q\upharpoonright\beta$.


Therefore $\pi$ is a projection.


