Many standard Algebraic Topology problems are to calculate fundamental groups by breaking the space apart into pieces you know the fundamental group of and then piecing these together. Most often this is done using the Seifert van Kampen Theorem. But I have never seen an example of how to compute a fundamental group of a less common space without this theorem. How does one compute the fundamental group when one of the conditions of the van Kampen Theorem fail?
I do not care about the spaces $U$, $V$ in the theorem being non-open - one really ought to make that demand if one is going to decompose the space. However, what do you do when the intersection is not path connected? There are many 'nice' spaces which have this property. For instance, I can easily sketch $3$ in Geogebra:
$1$. $X=U \cup V$, where $U$ is the sphere and $V$ is the torus - and $V$ passes through the sphere at two different circles where it enters/exists the torus. Now $X$, $U$, $V$ are all path connected. $U$ and $V$ are open. But $U \cap V$ is not path connected so the theorem fails.
$2.$ The same idea as in $(1)$ but instead we have two tori instead of a sphere and a torus. The issue with the van Kampen Theorem is the same
$3.$ $X=U \cup V$, where $U$ is a 'paper strip' and $V$ is the torus. The paper strip enters and exits the torus. Again, each is open and path connected but their intersection is four 'line segments' on the torus - certainly not path connected. The theorem fails again.
It is easy to think about the same and figure out what the fundamental group 'ought to be' - but I have no idea how one would actually prove it is the correct group. Is there a way for any of these examples - or even generally - that one should go about calculating the fundamental group when the intersection of $U$ and $V$ is not path connected?