A simple closed curve on a topological space $X$ is a continuous map $γ : [0, 1] → X$ such that $γ(0) = γ(1)$ and $γ|_{[0,1)} $ is one-to-one.
Show that the existence of a simple closed curve is a topological property.
I know that a topological property is defined to be a property that is preserved under a homeomorphism. Examples are connectedness and compactness.
So how do I show connectedness and compactness in here? Are there other properties?
Let $\psi:X\to Y$ be a homeomorphism and $\gamma: [0,1]\to X$ be a simple closed curve in $X$. Then, $\psi\circ \gamma:[0,1]\to Y$ is a simple closed curve in $Y$.
Note that $\psi\circ \gamma$ is continuous as the composition of two continuous functions is continuous.
Next, $\psi\circ \gamma(0)=\psi\big(\gamma(0)\big)=\psi\big(\gamma(1)\big)=\psi\circ \gamma(1)$.
Finally, $\gamma\big|_{[0,1)}$ is injective and $\psi$ is injective imply $\psi\circ \gamma\big|_{[0,1)}=\big(\psi\circ \gamma\big)\big|_{[0,1)}$ is injective.
Similarly, for any simple closed curve $\delta:[0,1]\to Y$ we can say $\psi^{-1}\circ \delta:[0,1]\to X$ is a simple closed curve in $X$.
What you are asked to show is that if $X$ is homeomorphic to $Y$ and there is a simple closed curve $\gamma $ in $X$ then there is simple closed curve in $Y$ too. Let $\phi:X \to Y$ be a homeom orphism. Define $\gamma '(t)=\phi (\gamma (t))$ for all $t \in [0,1]$. Verify that $\gamma '$ is a simple closed curve in $Y$.
