Let $X$ be a topological space and let $(X_i)_{i\in I}$ be a filtered family of subspaces. Let $X =\bigcup_{i \in I} X^°_i$ be the union of the interiors of the $X_i$. I want to prove the following statements:
- $\Pi_1(X) \cong \text{colim}_{i \in I} \Pi_1(Xi)$
- If $x \in \bigcap_{i \in I} X_i$, $\pi_1(X,x) \cong \text{colim}_{i \in I} \pi_1(X_i, x)$
So, to start, I want to show that $\Pi_1(X)$ satisfies the universal property of colimits in the category of groupoids. Hence, I consider a general groupoid $\Gamma$ with $h_i: \Pi_1(X_i) \to \Gamma, h_j: \Pi_1(X_j) \to \Gamma$, and look for the existence of a unique morphism $h: \Pi_1(X) \to \Gamma$ making the right triangles commute (where the map from $\Pi_1(X_i)$ to $\Pi_1(X)$ are just $\Pi_1(\iota_i)$, $\iota: X_i \to X$ being the inclusion, and similarly for $ \Pi_1(\iota_{ij}): \Pi_1(X_i) \to \Pi_1(X_j)$ for $i \leq j$).
Regarding the objects of $\Pi_1(X)$, these are just the points of $X$, thus if $x \in X$ then $x \in X_i$ for some $i \in I$, we are forced to set $h(x)=h_i(x)$ (I think?)
On morphisms: if I take a path $\gamma: I \to X$ in $X$, by compactness of $I$ I can show that $\gamma(I) \subseteq X_i$ for some $i \in I$, hence I can regard $\gamma$ as a path in $X_i$.
I would like to say that, if $h$ exists, then it must happen that $h([\gamma])=h_i([\gamma])$. In this case, $h$ would be uniquely determined. But what happens if $\gamma(I) \subseteq X_j \subseteq X_i$? Does it follow from the commutativity of the diagram that $h_j([\gamma])=h_i[\gamma])$? Can I use this as my definition of $h$?
Functoriality would come from the functoriality of $h_i$. Now, I know how to prove that if two loops are homotopic in $X$, then they are already homotopic in $X_k$ for some $k \in I$ (again by a compactness argument). Thus, if I take $\gamma, \alpha: I \to X$ such that they are homotopic in $X$, then from above we know that $\gamma: I \to X_i$ and $\alpha: I \to X_j$ for some $i, j \in I$, and from the filtered property we get that they are indeed homotopic in some $X_k, i,j \leq k$.
I would like to use this remark to show that $h$ depends only on the homotopy class of the path. The problem is that the I do not know how to incorporate this new $k \in I$ for which the two paths are homotopic in $X_k$, hence I am afraid that I am making some mistakes at the very beginning.
I can see that every morphism of $\Pi_1(X)$ and every pointed homotopy between such morphisms already lives in one of the subspaces $X_i$, but I don't understand how from these conditions I can get that $X$ is indeed the colimit.
I am completely lost now, thus any help will be highly appreciated. Thanks!