why $\pi_1(T) \cong \mathbb{Z}?$

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I have some confusion on this pdf(page no$:4$)

It is written that

compute $\pi_1(X) $ where $ X= \mathbb{R}^3 \setminus S^1$

take $U=X \setminus \mathbb{R}_z$ where $\mathbb{R}_z =\{(x,y,x)\in \mathbb{R}^3 :x=y=0\}$

$V= \{(x,y,x)\in \mathbb{R}^3 :x^2+y^2<1\}$

setting $T= \{(x,y,x)\in \mathbb{R}^3 :x>0,y=0\} \setminus\{(1,0,0)\}$

then $U $ is homeomorphic to $T \times S^1$

hence $\pi(U)\cong \pi_1(T) \times \pi_(S^1) \cong \mathbb{Z} \times \mathbb{Z}$

My confusion: why $\pi_1(T) \cong \mathbb{Z}?$

$T$ is not look like a circle .Then why $\pi_1(T) \cong \mathbb{Z}?$

Also,Im not able to visualize the diagram of $T$