Compute $\pi_1(S^1 \times (0,\infty))$

Question

Compute $\pi_1(S^1 \times (0,\infty))$

My attempt : By uisng the theorem $\pi_1( X\times Y) $ is isomorphic to $\pi_1(X) \times \pi_1(Y)$ if $X$ and $Y$ are connected

we have $\pi_1(S^1 \times (0,\infty))=\pi_1(S^1) \times \pi_1(0, \infty)$

Since $(0,\infty)$ is contractible so $\pi_1(0, \infty)=0$ and $\pi_1(S^1)=\mathbb{Z}$

$\implies \pi_1(S^1 \times (0,\infty))=\pi_1(S^1) \times \pi_1(0, \infty)=\mathbb{Z} \times \{0\}=0$

Therefore $\pi_1(S^1 \times (0,\infty))=0$

Is it true ?

Answer 1

So close. You made one mistake. In the category of groups, $A \times 0 \simeq 0$ does not hold. In fact, for groups, $0 = 1$, so $A \times 0 \simeq A \times 1 \simeq A$.

So $\mathbb{Z} \times 0 \simeq \mathbb{Z}$.