(Weak) homotopy type of mapping cones of inclusions

Question

For cofibrations $A \subset X$ the projection $CA \cup X \rightarrow X/A$ is a homotopy equivalence. But what happens for general pairs of spaces ? Is the projection at least still a weak equivalence or can the (weak) homotopy types actually differ ? I was not able to come up with any counterexample, but maybe someone here knows the answer.

Answer 1

Here is a kind of well known example.

Let $Y=A \vee B$ be the one-point union (wedge) of two Hawaian earings (look that up in wiki!), with base points the "bad point". Let $X=Y \cup CB$. Then $X/A$ is contractible. However $X \cup CA$ is a standard example of a space with an infinite fundamental group, with non homotopically trivial paths going alternately round the bases of the cones in ever decreasing lengths.

You can do this type of example another way round. Let $Z$ be the union of two (unreduced!) cones on Haiwaian earrings together with a line segment $L$ joining the bad points at the bases of the cones. Then $Z$ is contractible, as is $Z \cup CL$, but $Z/L\cong X \cup CA$ as above is not contractible.

There is a paper by Morgan and Morrison in Proc London Math Soc (3) (1986) 562-576 which has more detail on such (difficult) determinations of the fundamental group.