2
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ 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
2
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.