Let $B \leftarrow A \to C$ be a span of spaces, and consider the homotopy pushout $B \cup\_A C$.


**Question:** When is $B \cup\_A C$ contractible?


This is a pretty open-ended question. I'm interested in necessary conditions or sufficient conditions or interesting examples or special cases, etc.


**Some additional conditions I'd be happy to take as blanket assumptions:**


1. I don't think too much is lost if we assume that $A,B,C$ are connected.
2. I'm happy to assume that $A \to B$ is 1-connected (conventions vary about what this means; I mean that the homotopy fiber of $A \to B$ is 1-connected, i.e. that $A \to B$ is $\pi\_1$-surjective).
3. For somewhat obscure reasons, I'm particularly interested in the case where $A \to C$ is the projection $A\_0 \times C \to C$ out of a binary product.
4. For similarly obscure reasons, I'm particularly interested in the case where $C$ is a loopspace, or even where $C = \Omega \Sigma C\_0$ is a free loopspace.


**Observation:**


* By pasting on a few more pushout squares, we obtain the identity $\Sigma(B \vee C) = \Sigma A$.


I'm not sure what to make of this, though.



When is a homotopy pushout contractible?
Assume all the spaces are connected, $\pi\_1 A \to \pi\_1 B$ is surjective, and $\pi\_1 C$ is abelian. 


The pushout, $P = C \sqcup\_A B$, is contractible if and only if $\pi\_1(P)$ and all the $H\_i(P)$ are trivial. However, by Seifert-Van-Kampen, and the hypothesis on $\pi\_1 A \to \pi\_1 B$, we have that $\pi\_1 C$ surjects onto $\pi\_1 P$. Thus $\pi\_1 P$ is abelian, so $P$ is contractible if and only $H\_i(P) = 0$ for all $i$.


For this, as you and John observed, it is necessary and sufficent for $H\_i(A) \to H\_i(B) \oplus H\_i(C)$ to be an isomorphism for all $i \geq 1$. 


