Monotonic curvature and self intersections. I'm trying to prove that if $\alpha: I \to \Bbb R^2$ is a differentiable curve, where $I$ is an interval, has strictly monotonic curvature, then $\alpha$ has no self-intersections.
My attempt: We suppose WLOG that $\alpha$ is parametrized by arc-length. If the curvature $\kappa$ is strictly monotonic, and is defined on an interval, then $\kappa$ is injective. We consider $t_0, t_1 \in I$ such that $\alpha(t_0) = \alpha(t_1)$, and I want to prove that $t_0 = t_1$. If I manage to get $\kappa(t_0) = \kappa(t_1)$, then I'm done. For me, it is rather clear that this will be indeed the case, but I'm failing to justify that. My first thought was that: $$\alpha(t_0) = \alpha(t_1) \implies \alpha'(t_0) = \alpha'(t_1) \implies \alpha''(t_0) = \alpha''(t_1) \implies \kappa(t_0) = \kappa(t_1)$$
and so we would use $\kappa$'s injectivity to get $t_0 = t_1$. But the first two implications are clearly false, in general. Is there some way to use the hypothesis we have to conclude these implications, or is there another approach to the problem whatsoever? Thanks for your time.

On the other hand, I think that this might be false, since we could "glue" some curves together, like this: 

What to do?

Edit: As pointed by Semiclassical in the comments, the image on the pic is not a counter-example, since the curvature first increases, then decreases. But the fault in my first attempt still remains.

This is a corollary of the Kneser’s Nesting Theorem. Any self-intersection will give two osculating circles with at least one intersection, contradicting this theorem.
