Let $A,X, Y, I$ be topological spaces, $A$ being subspace of $X$ with $i$ being an inclusion, and $f:A \to Y$ being a continuous map. Let $(f,1): A \times I \to Y \times I$ denote map $(f,1)(a,t) = (f(a),t)$. Let $X \cup _{f} Y$ be an adjunction space.
(known, possibly well formulated) Theorem: Let $q : X \to X'$ be quotient map and $h:X \to Z$ such map that $q(x) = q(y)$ implies $h(x) = h(y)$. Then there exists unique map $h': X' \to Z$ such that $h' \circ q = h$.
(maybe wrong) Theorem: $X \cup _{f} Y \times I$ is homeomorphic to $(X \times I \cup _{(f,1)} Y\times I) $.
(maybe wrong) Proof:
Obviously $ (X \coprod Y) \times I \cong (X \times I) \coprod (Y \times I) $, let $F$ denote this homeomorphism.
Next, maps $(\text{pr},1):(X \coprod Y) \times I \to X \cup _{f} Y \times I$ and $pr' \circ F: (X \coprod Y) \times I \to (X \times I \cup _{(f,1)} Y\times I) $ are both quotient maps. (here $\text{pr}:(X \coprod Y)\to X \cup _{f} Y$ and $\text{pr}':(X \times I) \coprod (Y \times I) \to X \times I \cup _{(f,1)} Y\times I $ are canonical projections).
Applying the well known, possibly well formulated theorem to $q=(\text{pr},1)$ and $h = \text{pr}' \circ F$, we get that there is unique map $K:X \cup _{f} Y \times I \to X \times I \cup _{(f,1)} Y\times I$ and in similar manner there is unique map $K'$ going in opposite direction. $KK'$, $K'K$ must be identities, again my the same theorem. Q.E.D.
(maybe wrong) Corrolary: $X \cup _{f} Y \times I$ is pushout of maps $(i,1)$, $(f,1)$, i.e. functor $-\times I$ preserves adjunction spaces.
Where am I wrong please? Thanks in advance.