Suppose that $X$, $Y$ and $Z$ are topological spaces, with $A\subset X$, a map $f:A\rightarrow Y$, and a homotopy equivalence $\phi:Y\rightarrow Z$. It seems fair to think that the adjunction spaces $Y\cup_{f}X$ and $Z\cup_{\phi\circ f}X$ will be homotopy equivalent provided that $(X,A)$ has the homotopy extension property.
A reasonable candidate for a homotopy equivalence seems to arise from the map $(Id,\phi):X+Y\rightarrow X+Z$ after passing to the quotient ($X+Y$ denotes disjoint union, and $(Id,\phi)$ is the map defined to be the identity on $X$ and $\phi$ on Y).
Any suggestions will be appreciated.
Thanks!
Best Answer
Some cofibration assumptions are needed but the general result, a gluing theorem for homotopy equivalences, is about a map of pushout diagrams, one of which is as you give and the other is say $g: B \to Z$, and $ B \subset W$, in which say the vertical inclusion maps $ A \to X$, $B \to W$ are cofibrations. Then it says roughly: if the maps on $A,X,Y$ are homotopy equivalences, so is the map between the adjunction spaces.
This result appeared I think first in the 1968 edition of my book which is now Topology and Groupoids (2006), see 7.5.7 in that, and is a standard result in abstract homotopy theory. However the proof in the book has the advantage of giving closer control over the homotopies involved.
Many books on algebraic topology give invariants to show some spaces are not homotopy equivalent, but one also needs methods of proving directly that certain spaces are homotopy equivalent, and the gluing rule can be a convenient tool for this, as examples in the book show.
There is more detail and a cubical diagram on https://math.stackexchange.com/questions/184817