Show that a circle with a point removed have the same homotopy type with a point.
This is clear if we look at the picture, but I don't know how to actually write down a proof. I know that if space $X$ and $Y$ have the same homotopy type, then there exists continuous $f:X\to Y$ and $g:Y\to X$ such that $g(f) \sim id_X$ and $f(g)\sim id_Y$. So I am looking for these two maps. My first attempt was to map each point $(\cos\alpha,\sin\alpha)\to (0,0)$ and indeed this map is continuous. But then how can I find a map from $(0,0)$ to the circle? After all, how do we find a valid function from a single point to different points?