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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.

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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 $D_1\cup D_2$.

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May i ask something about this proof? Why is the connectedness property of $D_{1} \cap D_{2}$ important? Because otherwise $F(z)$ won't be well defined and holomorphic on the union of the simply connected sets? Thanks in advance! – Lullaby May 2 '13 at 9:22
@Lullaby: See my Addendum. – Christian Blatter May 13 '13 at 16:06

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.

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Am I right to assume that this is applied to the question with $\mathcal U=\{D_1,D_2\}$ and $x$ any point of $D_1\cap D_2$? – Marc van Leeuwen Oct 8 '12 at 12:00
@Marc: Exactly! – Ronnie Brown Oct 14 '12 at 16:33

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

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Thanks , but someone has an elementary proof? Not using notions of fundamental group , etc. ? That's what I want )= – Joseph Oct 6 '12 at 22:02
How do you intend to even define "simply connected" without talking about fundamental groups? – Chris Eagle Oct 6 '12 at 22:09
@ChrisEagle $\newcommand{\R}{\mathbb R}$ for the subsets of the plane, you can define an open subset $X \subset \R^2$ to be simply connected if $X$ is connected and $\hat{\R^2} \setminus X$ are connected. Where $\hat{\R^2}$ is the Riemann sphere. – JSchlather Oct 6 '12 at 22:17
I can define an homotopy between closed curves, and define an space to be simply connected, as a space such that all the closed curves are homotopic to a point – Joseph Oct 6 '12 at 22:35
This is a good question, as it shows a simple question can be best answered by a more sophisticated concept, in this case the fundamental group, or better, fundamental groupoid on a set of base points. – Ronnie Brown Oct 7 '12 at 10:25

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