Generalisation of Seifert-van Kampen theorem? The professor of the topology course stated the theorem, without proof, as:  

Seifert-van Kampen Theorem:
I. Let $X=\bigcup_{i=1}^nU_i,$ where $U_i$ are open and path-connected. Suppose $x_0\in \bigcap U_i$ and let $j_k:\pi_1(U_k,x_0)\rightarrow \pi_1(X,x_0)$ be induced maps of the natural inclusions $U_k\rightarrow X.$ Then these $j_k$ induce a unique group homomorphism $\Phi: \prod_{1\le i\le n}^*\pi_1(U_i,x_0)\rightarrow \pi_1(X,x_0)$ such that $\Phi(x)=j_k(x)$ if $x$ comes from $\pi_1(U_k,x_0),$ where the star on the product symbol indicates that this is the free product of these groups.
II. If $U_k\cap U_l$ is path-connected for all $k, l,$ then $\Phi$ is surjective.
  Further let $i_{k,l}:\pi_1(U_k\cap U_l,x_0)\rightarrow\pi_1(U_k,x_0)$ be induced by $U_k\cap U_l\rightarrow U_k.$
III. If $U_k\cap U_l\cap U_m$ is path-connected $\forall 1\le k, l, m\le n,$ then $\ker\Phi$ is the normal subgroup generated by all elements of the form $i_{k,l}(w)i_{l,k}(w)^{-1}, \forall w\in\pi_1(U_k\cap U_l,x_0).$  

But our textbook only states the case of the theorem where $n=2.$ And the proof in that case does not seem to apply in the general case: especially we only assume that the trifold intersections are path-connected, not including intersections of more than three opens.
I searched on the internet, and found nothing. I went to ask the professor, and he replied: it is a long story... We can talk about it in the course about algebraic topology during the next semester.
Apparently I don't think I can wait that long. So any hint or reference is sincerely welcomed. Thanks in advance.
 A: In answer to the request, here is the statement of the general Seifert-van Kampen Theorem for the fundamental groupoid on a set of base points, with the paper available here. The book Topology and Groupoids proves only the case of a union of two sets, and this using a retraction argument which does not apply easily to the general case, and not at all to higher dimensions. 
We first define for a space $X$ and set $A$ the fundamental groupoid $\pi_1(X,A)$ to be the set of homotopy classes rel end points of paths in $X$ joining points of $A \cap X$, with groupoid structure determined  by the usual composition of paths. We say $A$ is representative in $X$ if $A$ meets each path component of $X$. (Another term would be to say $(X,A)$ is connected.) 
Suppose $X$ has a cover $\mathcal  U = \{U_i: i \in I\}$  by open sets (or sets whose interiors cover $X$). Let $I^2$ be the set of pairs $\{(i,j): i,j \in I\}$. Let $U_{ij}= U_i \cap U_j $  and let $$a_{ij}: U_{ij} \to U_i, b_{ij} : U_{ij} \to U_j, c_i: U_i \to X$$ be the inclusions. For any set $A$ we have the diagram 
$$ \bigsqcup_{ij} \pi_1(U_{ij},A) \rightrightarrows^a_b \bigsqcup_i \pi_1(U_i,A) \to^c \pi_1(X,A), $$
where $a,b,c$ are induced by the above inclusions. Here $\bigsqcup$ denotes disjoint union, which is the coproduct in the category of groupoids. (In the category of groups, the coproduct is the free product, which needs more work to establish.) 
Theorem If the set $A$ is representative in all $1,2,3$-fold intersections of sets of $\mathcal U$, then the following holds: if $G$ is any groupoid and $f: \bigsqcup_{i} \pi_1(U_i ,A) \to G$ is any morphism of groupoids such that $fa=fb$, then there is a unique morphism $f': \pi_1(X,A) \to G$ such that $f'c=f$. 
Thus what is given is a universal property for $\pi_1(X,A)$, without a specific construction of something satisfying this universal property. The proof goes by verifying the universal property, which may   then be used to develop specific constructions. This is the categorical approach. 
