# Homotopy of closed curves is also a closed curve?

I'm trying to prove the following statement:

Let $\gamma_1$ and $\gamma_2$ be two closed curves from $[a, b]$ to $\mathbb{C}$, and let $h: [a,b]\times[0,1] \to \mathbb{C}$ be a homotopy between them. For any fixed $y_0 \in [0, 1]$, $h(x, y_0)$ is a closed curve $[a, b] \to \mathbb{C}$.

I'm not being able to make any progress though. Actually, I'm starting to suspect that it is not true, once I'm also failing to see why "opening" the closed curve and then closing it again fails to be a homotopy - it must fail, otherwise a loop with two windings around the origin could be homotopic to a loop with only one winding, which is not true!

PS: if it is actually false, I would be interested in the following weaker result, which may be simpler to prove:

Let $\gamma_1$ and $\gamma_2$ be two homotopic closed curves from $[a, b]$ to $\mathbb{C}$. Then, there exists a homotopy $h: [a, b] \times [0, 1] \to \mathbb{C}$ that satisfies: for any fixed $y_0 \in [0,1]$, $h(x, y_0)$ is a closed curve $[a, b] \to \mathbb{C}$.

That's all I need actually, to prove the FTA. But I also suspect that the first, stronger one, can be true.

• The first statement is indeed false. The second statement is true, but I'm not sure it's really what you mean to say--indeed, any two maps to $\mathbb{C}$ with the same domain are homotopic, since $\mathbb{C}$ is contractible. – Eric Wofsey Jul 14 '16 at 19:34
• @EricWofsey I'm trying to use this statement to prove that the winding number is invariant through homotopy. I've thought about showing that it is a continuous function on the $[0, 1]$ part of the domain that only assumes integer values, so it must be constant. It surely assumes integer values on $h(x, 0)$ and $h(x, 1)$, but I fail to see why this must be true for some value in between - i.e. why the homotopy can't open the curve and then close it again, as is exactly the "closed curve" property that forces it into an integer. – Henrique Augusto Souza Jul 14 '16 at 19:38
• If you're interested in winding number, then you presumably want the codomain of the homotopy to be $\mathbb{C}\setminus\{0\}$ or something, rather than $\mathbb{C}$. In any case, winding number is not homotopy-invariant unless you require each stage of the homotopy to be closed. – Eric Wofsey Jul 14 '16 at 19:40
• @EricWofsey I see. I hoped that wouldn't be the case (needing to force each stage to be closed), but it isn't a problem anyway. My closed curves are images of circles through polynomials, and the homotopy is given by the shrinking of the circle, therefore each stage will be naturally closed. Thanks for the help! – Henrique Augusto Souza Jul 14 '16 at 19:42