Path Connectedness in Van Kampen Theorem

Question

On page 17 of this pdf, http://www.math.uchicago.edu/~may/CONCISE/ConciseRevised.pdf, the Van Kampen Theorem is proven.

That is it is shown that for any covering of a space $X$ by a family of open path connected subsets $\lbrace U_i\rbrace$ closed under finite intersections, then the fundamental groupoid $\pi(X) =$ colim ${\pi(U_i)}$ Where the the category on $U$ is has the ${\pi(U_i)}$ as objects and inclusions $U_i \subset U_j$ as arrows.

Now the theorem is proven by showing that the required universal property is satisfied, that is that given any groupoid $G$ with family of maps $f_{U_i}:\pi(U_i) \to G$ such that the associated diagram commutes (that is that if $U_i \subset U_j$ we have $f_{U_j}|_{U_i} = f_{U_i}$) we can find a unique family of maps $f^*: \pi(X) \to G$ such that $f^*|_{\pi(U_I)} = f_{U_i}$.

For $a\in U_j$ we can define $f^*(a) = f_{U_j}(a)$. This is obviously unique and is well defined because if $a \in U_j, U_k$ we have $a \in {U_j \cap U_k}$ and $f_{U_j}|_{U_j \cap U_k} = f_{U_k}|_{U_j \cap U_k}$.

For a path $\phi: x \to y$ with $x, y \in X$, $f$ is covered by U and thus has a finite subcovering $\lbrace U_i\rbrace$. Now denote the restriction of our path to some path in $U_i$ $\phi_i$. It seems celar that we can find proper endpoints to do this. We can then define $f([\phi])$ to be the composite of all such $\phi_i$.

I think I can understand why the actual steps in the proof are valid, but the theorem seems to require that the sets involved be path connected despite that at no point the actual proof employs this. Can somebody please try to explain what part of that proof is invalid if the $\lbrace U_i\rbrace$ involved are not path connected? Why doesn't the map constructed still work?

Answer 1

I believe you are right in pointing out that the proof of the grupoid version of the theorem is still valid even after we drop path-connectedness assumptions.

However, in most algebraic textbooks Van Kampen theorem is stated in terms of fundamental groups and in fact this is also what Peter May does on the very next page. Here the path-connectedness is crucial, as one wants the fundamental grupoids of the open sets in the covering to be equivalent to fundamental groups (seen as categories). This is a possible explanation of this unnecessarily strong assumption given already in the grupoid version.

Answer 2

The general version of the Seifert-van Kampen theorem involves the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points, and for convenience we make this the classes of paths in $X$ joining points of $A \cap X$. Its statement is given in this stackexchange answer. It involves a cover $\mathcal U$ of $X$ by open sets, and the connectivity condition is that $A$ meets each path component of each $1,2,3$-fold intersection of sets of the cover. See also this paper for a proof which goes directly to the general case, and has the advantage of generalising to higher dimensions.

The answer to the question put is that if $A=X$ then that connectivity condition is automatically satisfied, and the proof for that case becomes very simple.

It is then quite a business to deduce the general case from this case, and Peter May deals only with the case $A$ is a single point, so avoiding a conclusion involving groupoids. My own view, and Grothendieck agreed, see this web page, is that the theorem is really about fundamental groupoids on a set of base points; I further hold that most of $1$-dimensional homotopy theory, including the above, covering spaces, and orbit spaces, are better expressed in terms of groupoids rather than groups. Peter goes some way in this direction for covering spaces.