I am trying to prove that the fundamental group of a wedge sum is actually the free product of the fundamental groups.
I read Hatcher's proof Algebraic Topology, but he omits some points / leaves them as implicit.
Here is my proof :
Let $\Omega$, $\Lambda$ be disjoint spaces and $x_0 \in \Omega, \; y_0 \in \Lambda$. Assume $x_0$ (resp. $y_0$) is the deform retract of an open neighborhood $U_{x_0}$ (resp. $U_{y_0}$) in $\Omega$ (resp. $\Lambda$). Let $A := \Omega \vee U_{y_0}$ and $B := \Lambda \vee U_{x_0}$. Since $\Omega$ and $\Lambda$ are disjoint, $A$ et $B$ are open, and besides $\Omega$ (resp. $\Lambda$) is retract deform of $A$ (resp. $B$) by the retract deform of $U_{y_0}$ into $\{y_0\} = \{x_0\}$ (resp. $U_{x_0}$ into $\{y_0\} = \{x_0\}$). And $A \cap B = U_{x_0} \vee U_{y_0}$ which deform retracts into the point $x_0$. Finally, $A \vee B$ deform retracts into $\Omega \vee \Lambda$. Hence, according to Seifert - Van Kampen theorem : $\pi_1(\Omega \vee \Lambda) = \pi_1(A \vee B) = \pi_1(A) \star_{\pi_1(A \cap B)} \pi_1(B) = \pi_1(\Omega) \star_{\{1\}} \pi_1(\Lambda) = \pi_1(\Omega) \star \pi_1(\Lambda)$.
Does it work ?
Especially I am not sure about the part "$A \vee B$ deform retracts into $\Omega \vee \Lambda$" : but without it I don't know how to conclude.
Thank you very much,
Respectfully,
AF
