I'm a student taking my first course in algebraic topology. I've stumbled across this exercise: calculate the fundamental group of $S^3-\gamma$, where $\gamma$ is a circumference in $\mathbb{R}^3$ (i.e. $\gamma=S^1$ in $\mathbb{R}^3$) and $S^3=\mathbb{R}^3\cup\lbrace\infty\rbrace$ is the one-point compactification of $\mathbb{R}^3$. I thought of $\mathbb{R^3}$ as $\mathbb{R}^1\times\mathbb{R}^2$, then I can think of $\gamma$ as an $S^1$ in $\mathbb{R}^2$ which has the same homotopy type as $\mathbb{R}^2\setminus\lbrace (0,0)\rbrace$. So $\mathbb{R^3}\cup\lbrace\infty\rbrace$ without $\gamma$ has the same homotopy type as $\mathbb{R}\cup\lbrace\infty\rbrace$, and its one-point compactification gives $S^1$.
In the end, I would get $\pi_1(S^3\setminus\gamma)=\pi_1(S^1)=\mathbb{Z}$, but I don't know if it's correct.
Maybe I could think of $S^3$ as $S^1\ast S^1$, but I don't know if that could help.
Thank you in advance