I wonder what is wrong in the following calculation of the fundamental group of the complement in $\mathbb R^3$ of the trefoil knot:
Blue lines are the boundaries of two open opposite overlaping semispaces, which divide the complement of the knot into two open sets $U$ and $V$. Then $U\cap V$ is homotopic to $\mathbb R^2$ minus four points, and hence $\pi_1(U\cap V)$ is a rank $4$ free group.
On the other hand, $U$ and $V$ are homeomorphic to open balls minus two lines joining two pair of points of the boundary, which are homotopic to open discs minus two points. So $\pi_1(U)$ and $\pi_1(V)$ are rank $2$ free groups.
Now, the map $\pi_1(U\cap V)\longrightarrow \pi_1(U)$ identifies the four generators in pairs, as well as $\pi_1(U\cap V)\longrightarrow \pi_1(U)$ does, but the identified pairs are different in each case.
Applying van Kampen's theorem, $\pi_1(U\cup V)$ is the amalgamated sum of $\pi_1(U)$ and $\pi_1(V)$, in which all four generators become identified. Hence $\pi_1(U\cup V)\cong \mathbb Z$.
This is false, of course, but, what is wrong in the above "proof"?