Showing that $\mathbb{C}$ minus a point is homotopy equivalent to $S^1$

Question

Intuitively this is clear (I think)? However I struggled to construct the homotopy. Here's what I attempted:

Define $M:=\mathbb{C} - p$ where $p$ is a point in $\mathbb{C}$.

Define also, $\alpha:S^1\rightarrow M$, given by $x \mapsto x$. And Define $\beta:M\rightarrow S^1$, given by $y\mapsto \frac{y}{\mid y \mid}$.

This gives us $\beta\alpha(x) = Id_{S^1}$. So all that is left to show is that and $\alpha\beta(x) = \frac{x}{\mid x \mid} \simeq Id_{M}$. Here's where I struggled to come up with a homotopy and perhaps that is because I picked $\alpha$ and $\beta$ wrong to start with?

Answer 1

Hint: Given two points $p,q \in \mathbb C$, how do you parameterize the line segment $\overline{pq}$ using the parameter interval $0 \le t \le 1$?