Suppose that $X$ is a CW-complex and $Y$ a CW-subcomplex of $X$. Let $A$ be a closed subspace of $Z$ such that 


1. $Z-A$ is homeomorhic to $X-Y$ and
2. $Z/A$ homeomorphic to $X/Y$ and
3. The closure of $Z-A$ in $Z$ is $Z$ it self.


Does it follow that the embedding $A\rightarrow Z$
is a cofibration ?



A pair of spaces equivalent to a pair of CW-complexes
If you mean cofibration in the sense of Quillen, then no. You can get a counterexample in which $Z-A$ is a $1$-cell.


Take $(X,Y)=(D^1,S^0)$. I will choose a compact space $Z$ with a dense open subset $U\cong D^1-S^0$. Then, writing $A=Z-U$, we get $Z/A$ is homeomorphic to $X/Y$, as they are both the one-point compactification of $D^1-S^0$.


$U$ will be a spiral in the open unit disk in the plane whose limit point set is the entire boundary circle. To be specific, map the unit complex disk to itself by $$re^{i\theta}\mapsto re^{i(\theta+\frac{1}{1-r})}$$Take the real open interval with endpoints $\pm 1$ and hit it with this map. 


Now if $Z$ is the closure of $U$ so that $A=Z-U$ is the unit circle in the plane then I claim $A\to Z$ is not a cofibration. I believe this follows from the fact that $A$ is one of the path-components of $Z$. Suppose for contradiction that $(Z,A)$ is a retract of a cellular pair $(W,B)$. Let $V\subset W$ be the union of all the path-components of $W$ having nonempty intersection with $B$. This is open and closed. Its preimage in $Z$ (under the coretraction $i:Z\to W$) is open and closed and contains $A$, so it is all of $Z$. But then a point $p\in U$ yields a point $i(p)\in V$ that can be joined by a path in $W$ to a point in $B$. Applying the retraction, we find that $p$ can be joined by a path in $Z$ to a point in $A$, contradiction.


