THis is problem 0.23 in Algebraic Topology of Hatcher
"Show that a CW complex $X$ is contractible if $X$ is the union of two contractible subcomplexes $A$ and $B$ whose intersection is also contractible."
We have a solution on this page CW complex is contractible if union of contractible subcomplexes with contractible intersection that says
"Obviously $A/A\cap B$ and $B/A\cap B$ cover $A\cup B/A\cap B$. Their intersection is a point. Indeed, take any point $p$ in their intersection. Take pre-image of $p$ under the quotient $A\cup B\to A\cup B/A\cap B$. It will be a subset of the intersection of pre-images, i.e. a subset of the $A\cup B$. Thus $p$ is the unique point in the intersection.
Since $A\cap B$ is contractible, the quotient $A/A\cap B$ is homotopy equivalent to $A$ itself. But $A$ is contractible. So you are done."
My question is, why do we need $A,B$ to be contractible? Clearly the solution here only uses the fact that $A\cap B$ is contractible. Did I overlook it?
Best Answer
Just to remove the question from the unanswered queue the contractibility of $A$ is used at the end of the proof.
To see why contractibility is needed think about $S^1\vee S^1$, the "figure 8" and take $A$ and $B$ to be the two circles. $A\cap B$ is a single point, hence contractible, but clearly $S^1\vee S^1$ isn't.