Spaces with fundamental group $\mathbb{Z}$

Question

Let $X=A\cup B$ be an open cover of $X$ where $A,B$ are simply connected and $A\cap B$ consists of 2 simply connected components $C_1, C_2$. Show that $\pi_1(X)=\mathbb{Z}$.

I tried different approaches similar to the proof that $\pi_1(S^1)=\mathbb{Z}$. The thing is I don't think I can find a simply connected covering because I don't think $X$ must be locally path connected. I did manage to show that every loop in $X$ is homotopic to an arbitrary loop starting at $C_1$, travelling through $A$ to a point in $C_2$, then returning to $C_1$ through $B$, repeating this "walk" some $n\in\mathbb{Z}$ number of times.

What I'm actually stuck with is showing that two such loops are not homotopic, particularly that one such walk around $X$ is not nullhomotopic. I thought assuming that this loop is nullhomotopic will imply that $A\cap B$ is connected, but I haven't found such a proof.

Answer 1

Have you tried a proof by contradiction using the Lebesgue number lemma? Assume that you have a loop of the form described, and assume that $H : [0,1] \times [0,1] \to X$ is a null homotopy of that loop. You get an open cover $H^{-1}(A) \cup H^{-1}(B) = [0,1] \times [0,1]$. Let $\lambda$ be a Lebesgue number. Subdivide the square $[0,1] \times [0,1]$ into little squares whose diameter is less than $\lambda$. Apply the Lebesgue number lemma. From there, see what conclusions you can draw.

Answer 2

This is surely a classic case for the version of the Seifert-van Kampen theorem involving the fundamental groupoid on a set of base points. I refer here to my answer to

https://mathoverflow.net/questions/40945/compelling-evidence-that-two-basepoints-are-better-than-one/46808#46808

and note that this groupoid version of the theorem was published in 1967, and developed in books, the current one being Topology and Groupoids. In the question, $A \cap B$ consists of $2$ simply connected components, so you need $2$ base points, and that gives rise to a groupoid. Here is a link to a proof of this groupoid version of the theorem. Note that the proof of the many pointed version is no harder than the proof of the single pointed version, but it just uses an extra and convenient concept, that of groupoid.

For more discussion see this discussion page on groupoids.

Here is an example of the situation of the question. Take two copies $I_1, I_2$ of the interval $[-1,1]$ and identify them at the corresponding points except for the $0$'s. The result is a non-Hausdorff space with fundamental group isomorphic to $\mathbb Z$.