Fundamental group with unit circle and one axis removed I am struggling with the following problem: Find the fundamental group of a space obtained by removing the unit circle from the x-y plane and z-axis from $\mathbb{R}^3$. I want to know what is the general strategy to solve problems like these. Any help will be greatly appreciated.
Regards,
And

First, check if the space is homotopy equivalent to a more familiar space.  In particular, see if the space deform retracts onto a simpler subspace. As BaronVT points out, $X=\mathbb{R}^3 \setminus (S^1 \cup \{\text{$z$-axis}\})$ deform retracts onto the (hollow) torus surrounding $S^1$. (If you're having trouble seeing this, try working with the space $Y=B^3 \setminus (S^1 \cup \{\text{$z$-axis}\})$, where $B^3 \subset \mathbb{R}^3$ is the unit ball; $X$ and $Y$ are homeomorphic.) It is often useful to imagine "pushing" the space away from whatever regions you've removed (say the $z$-axis) or "pulling" the space away from infinity (like in the deformation taking $\mathbb{R}^3$ to $B^3$).  
For many problems like this, you want to divide up the space into subspaces whose fundamental groups are more easily computed. Then you can use van Kampen's theorem to amalgamate them to get $\pi_1$ of the original space.
Instead of deforming the space or breaking it up into pieces, you can also try to take a different perspective on it.  For example, if you imagine adding a point at infinity to close up the $z$-axis of $\mathbb{R}^3$, you get $S^3$. If you remove this "closed $z$-axis" and the unit circle in the $xy$-plane from $S^3$, you get the complement of a Hopf link in $S^3$. Then you could compute the fundemental group of the space using techniques specific to this new perspective, e.g. the Wirtinger presentation.
