Fundamental group of $(\mathbb{R}\times\mathbb{R}^*)\cup\{(-1,0),(1,0)\}$.

Question

What is the fundamental group of $$X=(\mathbb{R}\times\mathbb{R}^*)\cup\{(-1,0),(1,0)\}=\{(x,y):y\not=0\}\cup \{(-1,0),(1,0)\}$$?

I'm guessing intuitively that the $X$ deformation retracts to $S^1$ but I'm not so sure.

EDIT: If I define $f:X\to S^1$ by $f(x)=\frac{x}{\|x\|}$, is this a retraction? If that is the case then $\pi_1(X)=\mathbb{Z}$ right?

Answer 1

Recall that $\mathbb R^2\setminus\{(0,0)\}$ deformation retracts to $S^1$: the map $F:[0,1]\times\mathbb R^2\setminus\{0\}\to S^1$ given by $$F(t,x)=\frac{x}{(1-t)+t\|x\|}$$ is such a deformation retraction.

Using this, can you build a deformation retraction from $X$ to $S^1$?