Consider two pointed CW complexes $(X,x_0),(Y,y_0)$ and their wedge sum $(X \vee Y,a)$ with $a$ the identification of the two base points. I want to give a sufficient condition under which we have
$$\pi_1(X \vee Y, a) = \pi_1(X,x_0) \ast \pi_1(Y,y_0) ~~.$$
Right now it is not clear to me whether we even need any additional assumption or whether this is true for any CW complexes. In case we need more conditions I am also looking for a counterexample to the case where there are no conditions.
One way I am considering is Seifert-Van-Kampen. Then we only need that there exists some neighborhood $U$ of $a \in X \vee Y$ which has trivial fundamental group. Sufficient for that would be:
Both, $X$,$Y$ have a neighborhood of $x_0$, $y_0$ which (stronlgy) deformation retracts to that point.
I am not sure whether this is true for CW complexes, but it could very well be. It is true that CW complexes are locally contractible, but that is not enough:
First off, for a contractible neighborhood $U$ of $x_0$ we do not even have $\pi_1(U,x) = 0$ (see comments). Moreover, there are contractible pointed spaces (even such that strongly deformation retract to a point) s.t. their wedge sum is not contractible. An example can be found here. However, this space is not a CW complex, so I do still have hope that the theorem holds without any further assumptions.
Best Answer
Yes, any point $x$ in a CW complex has a neighborhood which deformations retracts onto $x$. This is part of Proposition A.4 in Hatcher's Algebraic Topology for example.
Actually we do...? A contractible space has trivial homotopy groups. Is the issue with base points? A contractible space is path-connected, and any path between two points can be used to define isomorphisms between the fundamental groups based at these two points. So if a (sub)space $U$ deformation retracts onto some $x_0$, then $\pi_1(U,x) = 0$ for any $x \in U$.