# Example request: cofibrations need not induce cofibrations of adjunction spaces

The very first exercise in May and Ponto's More Concise Algebraic Topology asks for a demonstration of the following claim: given a commutative diagram $$\require{AMScd} \begin{CD} Y @<<< X @>i>> Z\\ @VVV @VVV @VVV \\ Y' @<<< X' @>i'>> Z' \end{CD}$$ of spaces in which the vertical maps are cofibrations and $i$ and $i'$ closed inclusions, it is not the case that the induced map $$Y \cup_X Z \longrightarrow Y' \cup_{X'} Z'$$ need also be a cofibration. They claim in fact there is an example in which $i'$ is an equality. What is such an example?

While I'm at it, the second, dual question asks for a demonstration that given $$\require{AMScd} \begin{CD} Y @>>> X @<<< Z\\ @VVV @VVV @VVV \\ Y' @>>> X' @<<< Z', \end{CD}$$ where the vertical maps are now fibrations, it need not be true the induced map $$Y \times_X Z \longrightarrow Y' \times_{X'} Z'$$ is a fibration. Again, why?

Let $X$ be a nonempty CW complex. Consider the diagram $$\begin{array}{ccccc} X & \leftarrow & \emptyset & \rightarrow & X \\ \downarrow & & \downarrow & & \downarrow \\ X & \leftarrow & X & \rightarrow & X \end{array}$$ where all the arrows $X \to X$ are the identity. The map of pushouts is the fold map $X \sqcup X \to X$, which is not injective, so not a cofibration.
We can more or less dualize this to obtain the second counterexample. Let $Y$ be a path-connected space with more than two points, and consider $$\begin{array}{ccccc} Y & \rightarrow & Y & \leftarrow & Y \\ \downarrow & & \downarrow & & \downarrow \\ Y & \rightarrow & * & \leftarrow & Y. \end{array}$$ Then the map of pullbacks is the diagonal map $Y \to Y \times Y$, which is not a fibration.