Let $\left(X,A\right)$ be a cofibered pair. Has pair $\left(X\times\left\{ t\right\} ,A\times\mathbb{I}\right)$ the gluing property? Let $\left(X,A\right)$ be a cofibered pair. Then: 


*

*$X\times\left\{ 0\right\} \cup A\times\mathbb{I}$ is a retract of
$X\times\mathbb{I}$.

*Pair $\left(X\times\left\{ 0\right\} ,A\times\mathbb{I}\right)$ has
the gluing property.
Here 2. means that the following diagram (with inclusions) is a pushout square: 
\begin{matrix}A\times\left\{ 0\right\}  & \rightarrow & A\times\mathbb{I}\\
\downarrow &  & \downarrow\\
X\times\left\{ 0\right\}  & \rightarrow & X\times\left\{ 0\right\} \cup A\times\mathbb{I}\end{matrix} It can be shown that 1. implies 2. but that is not the issue here. Fix $t\in\mathbb{I}$.
On base of the first statement a retraction of $X\times\mathbb{I}$
onto $X\times\left\{ t\right\} \cup A\times\mathbb{I}$ can be constructed.
My question is: can it also be shown here that pair $\left(X\times\left\{ t\right\} ,A\times\mathbb{I}\right)$
has the gluing property? Underlying is the question: if $\left(X,A\right)$
is a cofibered pair and the following diagram is a pushout square
with $Q=B\cup\bar{F}\left(X\times\mathbb{I}\right)$ :
\begin{matrix}A\times\mathbb{I} & \stackrel{F}{\rightarrow} & B\\
\downarrow &  & \downarrow\\
X\times\mathbb{I} & \stackrel{\bar{F}}{\rightarrow} & Q\end{matrix}
can it be concluded that the following diagram with $Q_{t}=B\cup\bar{f}_{t}\left(X\right)$
is a pushout square? 
\begin{matrix}A & \stackrel{f_{t}}{\rightarrow} & B\\
\downarrow &  & \downarrow\\
X & \stackrel{\bar{f}_{t}}{\rightarrow} & Q_{t}\end{matrix}
 Maybe I am looking in a wrong direction and is there a simple answer
to this underlying question.
 A: The answer is yes. Arne Strøm proved this to be true for $t=0$ in Note on Cofibrations II. I thought that $t=0,1$ were special cases here, but if I have not overlooked anything then actually they are not. 

Let $X$ be a topological space and let $A\subset X$. Fix some $t\in\mathbb{I}$
and define $Y_{t}=X\times\left\{ t\right\} \cup A\times\mathbb{I}$.
Let $r:X\times I\rightarrow Y_{t}$ be a retraction. Go out from inclusion
$i:Y_{t}\rightarrow X\times\mathbb{I}$ and projections $p:X\times\mathbb{I}\rightarrow X$
and $q:X\times\mathbb{I}\rightarrow\mathbb{I}$. Define $r_{1}=pir:X\times\mathbb{I}\rightarrow X$
and $r_{2}=qir:X\times\mathbb{I}\rightarrow\mathbb{I}$. 
It is our
aim to show that the pair of sets $\left(X\times\left\{ t\right\} ,A\times\mathbb{I}\right)$
has the gluing property. For this assume that $O\subset Y_{t}$ with
$X\times\left\{ t\right\} \cap O$ is open in $X\times\left\{ t\right\} $
and $A\times\mathbb{I}\cap O$ is open in $A\times\mathbb{I}$. Then
$X\times\left\{ t\right\} \cap O=W\times\left\{ t\right\} $ for some
$W$ open in $X$ and $A\times\mathbb{I}\cap O=A\times\mathbb{I}\cap V$
for some $V$ is open in $X\times\mathbb{I}$. To be shown is that
$O$ is open in $Y_{t}$. It is sufficient to find for every $\left(x,s\right)\in Y_{t}\cap O$
an open $N_{\left(x,s\right)}$ in $X\times\mathbb{I}$ with $\left(x,s\right)\in N_{\left(x,s\right)}$
and $Y_{t}\cap N_{\left(x,s\right)}\subset O$. Then $N=\cup\left\{ N_{\left(x,s\right)}\mid\left(x,s\right)\in Y_{t}\cap O\right\} $
is an open set in $X\times\mathbb{I}$ with $O=Y_{t}\cap N$, showing
that $O$ is open in $Y_{t}$. If $s\neq t$ then $N_{\left(x,s\right)}=X\times\left(\mathbb{I}\backslash\left\{ t\right\} \right)\cap V$
suffices. If $s=t$ and $x\notin\overline{A}$ then $N_{\left(x,t\right)}=\left(W\backslash\overline{A}\right)\times\mathbb{I}$
suffices. From here we concentrate on elements of $\overline{A}\cap W\times\left\{ t\right\} $.
Define $U\subset X$ by stating that:

$x\in U$ if open sets $P$ and
  $Q$ exists with $\left(x,t\right)\in P\times Q$ and $\left(A\cap P\right)\times Q\subset O$.

If $x\in U$ then $N_{\left(x,t\right)}=\left(P\cap W\right)\times Q$
suffices, so we are ready if has been shown that $\overline{A}\cap W\subset U$.
We have $\left(A\cap W\right)\times\left\{ t\right\} \subset V$.
For $x\in A\cap W$ find open sets $P$ and $Q$ open with $\left(x,t\right)\in P\times Q\subset V$.
Then $\left(A\cap P\right)\times Q\subset O$ so $x\in U$. This proves
that $A\cap W\subset U$. Now let $x\in\overline{A}\cap W$. We have
$\left(x,t\right)\in r_{1}^{-1}\left(W\right)$ and this set is open.
Consequently some $s\in\mathbb{I}\backslash\left\{ t\right\} $ exists
with $r_{1}\left(x,s\right)\in W$. Set $r_{2}^{-1}\left(\left\{ s\right\} \right)$
is closed so from $A\times\left\{ s\right\} \subset r_{2}^{-1}\left(\left\{ s\right\} \right)$
it follows that $\overline{A}\times\left\{ s\right\} =\overline{A}\times\overline{\left\{ s\right\} }=\overline{A\times\left\{ s\right\} }\subset r_{2}^{-1}\left(\left\{ s\right\} \right)$.
Then $r\left(\overline{A}\times\left\{ s\right\} \right)\subset Y_{t}\cap X\times\left\{ s\right\} =A\times\left\{ s\right\} $
so $r_{1}\left(x,s\right)\in A$. Then $r_{1}\left(x,s\right)\in U$
since $A\cap W\subset U$. Let $P$ and $Q$ be open sets with $\left(r_{1}\left(x,s\right),t\right)\in P\times Q$
and $\left(A\cap P\right)\times Q\subset O$. Then $\left(x,s\right)\in r_{1}^{-1}\left(P\right)$
and this set is open. Consequently some open $R$ exists with $x\in R$
and $r_{1}\left(R\times\left\{ s\right\} \right)\subset P$. Then
$A\cap R\subset r_{1}\left(R\times\left\{ s\right\} \right)\subset P$
so that $\left(A\cap R\right)\times Q\subset\left(A\cap P\right)\times Q\subset O$.
This allows the final conclusion that $x\in U$.
