union of two simply connected open , with open and non empty intersection in $R^2$

Question

Let $D_1,D_2$ be two simply connected open subsets of $\mathbb{R}^2$. Let's suppose that it's intersection is nonempty and connected. Then $ D_1\cup D_2$ is simply connected. I have no idea how can I do this.

Answer 1

If you want a result without groups or groupoids, here it is.

Let $\cal U$ be an open cover of a space $X$ with base point $x$ such that each set of $\cal U$ is simply connected and each set of $\cal U$ and intersection of two sets of $\cal U$ is path connected and contains $x$. Then $X$ is simply connected.

Proof. Let $a: I \to X $ be a loop in $X $ at the base point $x $. By the Lebesgue covering lemma, there is a subdivision $ a=a_1 + a_2+ \cdots +a_n $ of $a $ such that each $a_i $ lies in some set $U_i $ of $\cal U$. We consider $a_i$ as a map $I \to X$. Let $b_0, b_{n+1}$ be constant paths at the point $x$. By the assumptions, we can for $1 < i \leq n$ choose a path $b_i: I \to X$ joining $a_i(0)$ to $x$ and lying in $U_{i-1} \cap U_i$. The loop $-b_i + a_i+b_{i+1}$ lies in $U_i$ and so is contractible to a constant rel end points, by assumption of simple connectivity of $U_i$. It follows that $a$ is contractible to a point in $X$.

This argument can be usefully generalised to higher dimensions, in the context of filtered spaces.

Answer 2

The following argument uses a "known fact" from complex analysis, namely: A domain $\Omega\subset{\mathbb C}$ is simply connected iff for any $a\notin \Omega$ the function $f_a(z):={1\over z-a}$ has a primitive $F:\ \Omega\to{\mathbb C}$ . (Intuitively: If $\Omega$ has no holes were a "bad" $a$ can be placed.)

Fix an $a\notin D_1\cup D_2$. Then $f_a\restriction D_i$ has primitives $F_i:\ D_i\to{\mathbb C}$, since the $D_i$ are simply connected. The function $\phi:\ z\mapsto F_1(z)-F_2(z)$ is well defined on $D_1\cap D_2$ and has zero derivative. Since $D_1\cap D_2$ is connected we conclude that $\phi(z)\equiv c$ for some $c\in{\mathbb C}$. Then $$F(z):=\cases{F_1(z) &$\quad(z\in D_1)$\cr F_2(z)+c&$\quad(z\in D_2)$\cr}$$ is a well defined primitive of $f_a$ on $D_1\cup D_2$.

As $a\notin D_1\cup D_2$ was arbitrary, the conclusion follows.

Addendum: The connectedness of $D_1\cap D_2$ was essential here. Without it we only can say that $\phi$ is a constant $c_\alpha$ on each component $C_\alpha$ of $D_1\cap D_2$. As a consequence there is no obvious way to define the above $F$ on $D_1\cup D_2$, and in most cases there is no such $F$ at all. An example: Let $$D_1:=\Bigl\{z\in{\mathbb C}\>\Bigm|\> z\ne0,\ |{\rm Arg}\> z|<{2\pi\over3}\Bigr\},\quad D_2:=\{-z\>|\>z\in D_1\}\ .$$ Here $D_1\cap D_2$ is not connected, but all the other assumptions are fulfilled. And sure enough, the union $D_1\cup D_2$ is not simply connected, nor does the function $f(z):={1\over z}$ have a primitive on $D_1\cup D_2$.

Answer 3

This is a small special case of van Kampen's theorem.