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 $\Omega$.


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?

  • $\begingroup$ Isn't $(s, t) \mapsto F(\gamma(s), t)$ the homotopy that you are looking for? Or am I misunderstanding something? $\endgroup$
    – Martin R
    Apr 24 '20 at 11:10
  • $\begingroup$ @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). $\endgroup$
    – Zuy
    Apr 24 '20 at 11:16
  • $\begingroup$ OK, I see what you mean. $\endgroup$
    – Martin R
    Apr 24 '20 at 11:18
  • $\begingroup$ Your homotopy $F$ is backward?! $\endgroup$
    – C.F.G
    Apr 24 '20 at 15:40
  • $\begingroup$ 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. $\endgroup$ Apr 24 '20 at 17:52

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

  • $\begingroup$ Thank you. This works. How can I show this equivalence? $\endgroup$
    – Zuy
    Apr 24 '20 at 12:46
  • $\begingroup$ 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). $\endgroup$
    – Lee Mosher
    Apr 24 '20 at 12:57
  • $\begingroup$ I'll add one more thing: the major ideas behind all of this proof involve quotient maps; true "algebraic topology" is not needed. $\endgroup$
    – Lee Mosher
    Apr 24 '20 at 12:59

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