Intereting Posts

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the solution for an integral including exponential integral function
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The class of equivalence.

While reading Markushevich’s complex analysis book, I realized that his definition of a simply connected domain differs from the one I have seen before. He takes the Jordan Curve Theorem for granted, and denotes the interior of a closed Jordan curve by $I(\gamma)$. Then he defines (pages 70~72 of vol.1) ;

A domain $G$ is simply connected iff whenever $G$ contains a closed Jordan curve $\gamma$, $G$ also contains $I(\gamma)$.

Then he shows some interesting results, such as a bounded domain is simply connected if and only if its boundary is connected (of course, with his definition).

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- Geometric interpretation of reduction of structure group to $SU(n)$.
- Star-shaped domain whose closure is not homeomorphic to $B^n$
- Examples of group extension $G/N=Q$ with continuous $G$ and $Q$, but finite $N$

My question is : **Does the above definition coincides with the usual one, given by the condition $\pi_1(G)=1$ for the fundamental group of $G$?**

I know that this has to be the case since otherwise the whole theory would break down.. but how exactly can we explain this equivalence? Do we need the Jordanâ€“Schoenflies theorem?

Any advice is welcome. Any reference is also welcome.

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- Contour Integral: $\int^{\infty}_{0}(1+z^n)^{-1}dz$
- How to prove that the Weierstrass $\wp$-function is a well defined meromorphic function on a torus?
- Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$

Here’s a rough idea as to why you might expect these definition to be the same. You might run into more problems making this precise, but this should hopefully give some intuition.

Suppose we have a point in $G$ but not the interior of $\gamma$. If $\pi_1(G)=1$ then we can contract $\gamma$ to this point. But then it very much looks like the point $x$ is in the interior of $\gamma$! (This should rely on the fact that the curve is simple)

The other direction seems a bit more difficult, since the curves we consider need not be simple. But for any Jordan curve, using the property that its interior is contained in $G$, we can use “straight line homotopy” to map contract it to a point $x\in I(\gamma)$ (map every point along a straight line to the point $x$). Then in the general case you could argue in two ways. The first would be to somehow use $\pi_1(S^1) \cong \mathbb{Z}$ to say that the curve should be homotopic to a Jordan curve. The second could be to consider each bounded connected component of the curve. Each is bounded by a Jordan curve, so the same argument would say that there is no puncture or anything in this component.

In any case, thinking about how the alternative definition of simply connected rules out non-contractible loops should get you most of the way there to understanding this. I would recommend drawing lots of pictures!

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