Annulus Homotopic to punctured plane I know a circle is homtopic to a punctured plane, and by the same reasoning, the aannulus must also be, as it a "step" in the homotopy (IE the annulus is a "stretched" circle). The only trouble is it is not clear how to construct the required $f,g$. Any thoughts?

Fact, $S^1\simeq \mathbb{R}\setminus \\{(0,0)\\}$. I want to show that $Y=\\{(x,y)\in \mathbb{R}^2| 1\leq x^2+y^2 \leq 2\\} $ is homotpic to $\mathbb{R}\setminus \\{(0,0)\\}$. This can be done by showing $Y\simeq S^1$. Let $f:S^1\to Y$ send $(x,y)\to (x,y)$. For $(x,y)=z$, let $g(z)= \frac{z}{||z||}$. Clearly, $g\circ f$ is the identity on $S^1$ and is homotopic to the identity. For $f\circ g$ , define $F: Y\times I \to A $ by $F((z,t)= \frac{t||z||+ (1-t)}{||z||}z$ This map is continuous being the composition of continuous maps, and $F((x,y),0)=(f\circ g)(x,y)$ and $F((x,y),1)=(x,y)$. Hence, $f$ and $g$ are a homotopy equivilance from $Y$ to $S^1$. Therefore, since $Y\simeq S^1$, $\implies y\simeq \mathbb{R}\setminus \\{(0,0)\\}$.