# Simply connected and homotopic

In complex plane, if $C$ is a closed curve that is homotopic to a point, and $C$ is the boundary of a domain $E$, is $E$ simply connected?

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You need to clarify few things in your question: 1. Homotopic to a point in which space? (In the plane every loop is homotopic to a point!) 2. What do you mean by "bounds a domain"? For instance, does unit circle bound the complement to the closed unit disk? 3. Is the curve supposed to be simple? 4. Is the domain assumed to be connected? – studiosus Mar 10 '14 at 19:43

I think you need to assume that $C$ is a simple closed curve. Otherwise, take a long skinny ellipse (a "sausage") and bend it until the two ends just touch, like your thumb touching your forefinger when you use them to make a circle. There's clearly a curve that bounds the sausage (even after the ends touch at a single point), and this curve is clearly contractible within the sausage shape, but the sausage shape (assuming it includes the boundary) is not contractible -- it has $\pi_1 = \mathbb Z$.

Even with the simple-closed-curve assumption I'm not entirely convinced the statement is true, but I suspect that it is.

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>the sausage shape (assuming it includes the boundary) is not contractible< Do you mean the sausage to be $E$? Well $E$ is supposed to be a domain so must be open. – eltonjohn Mar 6 '14 at 8:41
I didn't realize that in the context of complex variables, "domain" always meant "open." I probably once know that, a long time ago. Anyhow, point taken. – John Hughes Mar 10 '14 at 19:37

I have to rewrite the whole what I said:

I think you need some conditions on $C$ and $E$ to avoid pathology and to say something useful. The conditions that come to my mind include:

1) $E$ is path-connected, (since $E$ is a domain, it is open and connected but I don't think they are strong enough to prevent pathological phenomena to take place.)

2) the pair ($E, C$) is reasonably tame, for example every point of $C$ has a neighbourhood in $E \cup C$ which is homeomorphic to $(-1, 1) \times [0, 1)$ where (0, 0) corresponds to the point,

3) the homotopy of $C$ to a point occurs inside $E$.

Of course you may need more.

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shrink to what point? – 89085731 Mar 6 '14 at 3:31
You wrote "C is a closed curve that is homotopic to a point", didn't you? – eltonjohn Mar 6 '14 at 3:37
On second thought if you meant, by `$C$ is homotopic to a point', that $C$ is shrinkable to a point of $E$ inside $E$, then that homotopy can be used to define a contraction of $E$ to the point inside $E$, thus $E$ is simply connected. – eltonjohn Mar 6 '14 at 4:06
The quick way I see to use the homotopy to define a contraction would be something like "For each point $P$, find the first moment $H(s_0, t)$ where $H(s_0, t) = P$, for some $t$, and then let $P$ follow the path $H([s_0, 1], t)$." But that assumes that there is such a first moment, and that $(s_0, t)$ depends nicely on $P$. I see lots of proving left to do. Or did you have something else in mind that's simpler? – John Hughes Mar 6 '14 at 12:20
>I see lots of proving left to do.< I quite agree. Besides, as I rewrote, we need many postulates to bypass pathology. So, this problem is not so easy as it appears. – eltonjohn Mar 6 '14 at 15:20