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?
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