Homotopy between any loop in an annulus and a loop in a circle contained in the annulus

Question

Show that any loop in the annulus $\textbf{A}=\{(x,y) \big | 1 \le x^2 + y^2 \le 9 \} \subset \mathbb{R}^2$ with base point $(2,0)$ is homotopic to a loop whose image lies in the circle $A=\{(x,y) \big | x^2+y^2=4\}$.

I am not sure if I am understanding this question correctly. Is it not true that any loop in the circle with base point at $(0,2)$ must be the circle, $A$, itself?

Furthermore, in the annulus I can see two types of loops. Those that transverse the hole, and those that are homotopic to the constant loop.

Are my assumptions correct?

Answer 1

No, there are loops in the circle whose image is not the full circle (for instance the constant loop).

Up to homotopy, there are a copy of $\mathbb{Z}$ types of loops in the annulus, given by the winding number of the loop around the hole (if you have not proved this yet in your class, then you soon will when you discuss the fundamental group of a topological space).

As far as the question in the title goes, consider the map from the annulus to the circle which maps a point $(x,y)$ to $\frac{(1-t)|(x,y)|+2t}{|(x,y)|}(x,y)$ for $t\in[0,1]$.

Answer 2

It sounds like you are making an invalid assumption about the concept of a "loop". The definition of a "loop in $X$ with base point $p$" is a continuous function $[0,1] \to X$ such that $0 \to p$ and $1 \to p$.

So, for example, the statement "any loop in the circle $A$ with base point at $(0,2)$ must be the circle $A$ itself" is nonsense. A loop in $X$ is not a subset of $X$, it is a certain kind of function with values in $X$.

Perhaps you meant to say "any loop in the circle $A$ with base point at $(0,2)$ must have image equal to the circle $A$ itself"? That statement at least makes sense, but it is false: take the constant loop.