# Show that a circle with a point removed has the same homotopy type with a point.

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?

Regard the circle as the quotient space $S^1=[-1,1]/_\sim$ where $x\sim y$ iff $x=y$ or $\{x,y\}=\{-1,1\}$. Now we may choose to remove the point $-1\sim1$ from the space as any other choice of removed point can be obtained by a homeomorphism. Now $S^1\setminus\{q\}\cong(-1,1)$. Now we have reduced the problem to showing $(-1,1)$ is homotopic to a point. Define $f:(-1,1)\to\{p\}$ and $g:\{p\}\to(-1,1):p\mapsto0$. Immediately $f\circ g=id_{\{p\}}$. Define $H:(-1,1)\times[0,1]\to(-1,1):(x,t)\mapsto xt$. This is a continuous function, and indeed a homotopy of the functions $(g\circ f)(x)=H(x,0)$ and $id_{(-1,1)}(x)=H(x,1)$. Hence $S^1\setminus\{q\}\simeq\{p\}$ as desired.