# Contractible implies simply connected

Question:

Let $$\Omega\stackrel{\text{open}}{\subseteq}\mathbb{C}$$ be contractible, i.e. there exist $$z_0 \in \Omega$$ and a continuous map $$F:\Omega \times[0,1]\to \Omega$$ satisfying $$\forall z \in \Omega: F(z,0)=z_0 \text{ and } F(z,1)=z.$$

Moreover, let $$\gamma:[a,b]\to \Omega$$ be a closed curve satisfying $$\gamma(a)=z_0=\gamma(b)$$.

Show that $$\gamma$$ is null-homotopic in $$\Omega$$.

Comments:

My goal is actually to show that a contractible open subset of the complex plane is simply connected. Above claim is equivalent to this.

My problem is that in general, we do not have $$F(z_0,t)=z_0$$ for all $$t\in[0,1]$$. Otherwise it would be easy to construct a homotopy transforming $$\gamma$$ into a constant curve.

I have not taken a course about algebraic topology (AT) yet. Hence I also have no equivalent definition of "contractible" yet. I know that similar questions have been asked before, but all of them applied results of AT.

Any ideas to solve this without results of AT?

• Isn't $(s, t) \mapsto F(\gamma(s), t)$ the homotopy that you are looking for? Or am I misunderstanding something? Commented Apr 24, 2020 at 11:10
• @MartinR Thank you for your comment. This was my idea as well. But in general, this is not a homotopy: A homotopy $h$ from $\gamma$ to $z_0$ must satisfy: $\forall t \in [0,1]: h(a,t)=z_0$. Your map doesn't ensure this (see my second comment).
– Zuy
Commented Apr 24, 2020 at 11:16
• OK, I see what you mean. Commented Apr 24, 2020 at 11:18
• Your homotopy $F$ is backward?! Commented Apr 24, 2020 at 15:40
• Yes, you need to do some basepoint conjugation tricks to deal with the basepoint. It's not a big deal to assume the homotopy is based to simplify the idea for now, though, imo.
– user17892
Commented Apr 24, 2020 at 17:52

## 1 Answer

Hint: Use an alternate (and equivalent) characterization of simple connectivity: $$\Omega$$ is simply connected if and only if it is path connected and every continuous function $$f : S^1 \to \Omega$$ is homotopic to a constant function.

Second hint: To prove the operative direction of the equivalence: think of a given closed curve $$\gamma : [0,1] \to \Omega$$ as a function $$\Gamma : S^1 \to \Omega$$ (using the quotient map $$[0,1] \to S^1$$ defined by $$t \mapsto (\cos(2\pi t),\sin(2\pi t))$$; assume the existence of a homotopy $$H : S^1 \times [0,1] \to \Omega$$ from the function $$H(x,0)=\Gamma(s)$$ to a constant function $$H(x,1) = z$$; consider the point $$1 \in S^1$$ which satisfies $$H(1,0)=z_0$$ and $$H(1,1)=z$$; consider the path $$H(1,t)$$ between $$z_0$$ and $$z$$; then stare at all that information and use it to construct a homotopy from $$\gamma$$ to a constant path at $$z_0$$ such that which $$\gamma(0)=\gamma(1)=z_0$$ does not move under the homotopy

• Thank you. This works. How can I show this equivalence?
– Zuy
Commented Apr 24, 2020 at 12:46
• I added another hint. I'll add also that the proof of that equivalence is an exercise in Hatchers book "Algebraic Topology"; and I believe that it is also an exercise, or perhaps even a proved statement, in Munkres "Topology" (but I don't have my copy with me to check). Commented Apr 24, 2020 at 12:57
• I'll add one more thing: the major ideas behind all of this proof involve quotient maps; true "algebraic topology" is not needed. Commented Apr 24, 2020 at 12:59