Let $A$ and $B$ be two simplicial complexes (or CW-complexes) containing a common subcomplex C. Assume that C is contractible in both A and B.
Is it true that the space obtained gluing A and B over C homotopy equivalent to $$A \vee B \vee \Sigma C$$ where $\Sigma C$ is the suspension of C ?
Do you have any idea about that ?
Thanks in advance.
This seems false. Suppose that $A$, $B$, and $C$ are all the same triangle (i.e., 3 vertices, 3 edges). Then so is the glue-up space, $X$. But $\Sigma C$ is a topological 2-sphere, so it has nontrivial second homology, while $X$ does not.
To make the example less trivial, you can take $A$ and $B$ to be $C \times I$, where $C \times \{1\}$ is the embedding of $C$ in $A$, and $C \times \{0\}$ is the embedding of $C$ in $B$, so that the space $X$ just looks like $C \times [0, 2]$.