Complement of a plane in $\mathbb{R}^4$ I have problems understanding these two spaces:
1) the complement to a plane in $\mathbb{R}^4$
2) the complement to the circle $S^1$ in $S^3$.
what are they homotopic to?
 A: Think of the plane in $\mathbb R^4$ as being spanned by the first two coordinate axes: $\{(x,y,0,0):x,y\in\mathbb R\}$. Then projection onto the second two coordinates is a homotopy equivalence $(x,y,z,t)\mapsto (z,t)$ onto $\mathbb R^2\setminus\{(0,0)\}$. Now $\mathbb R^2\setminus\{(0,0)\}$ is homotopy equivalent to a circle.
For $S^1$ in $S^3$, recall that $S^3$ is a union of two solid tori glued along their boundary in a certain way. Think of your circle $S^1$ as being the core of one of these tori. Then removing it will be homotopy equivalent to the other solid torus, which is homotopy equivalent to a circle $S^1$. If your circle sits in $S^3$ as a knot, then this won't work, and in fact the homotopy type of knot complements is a very rich set.
A: To add to the other two answers, it is also worthwhile to see that the homotopy of $\mathbb{R}^4$ minus a plane to $S^1$ can be made to factor through $S^3$ minus the unknot. That is, $\mathbb{R}^4 \setminus \{(x,y,0,0)\}$ deformation retracts to $\{(x,y,z,w)\mid x^2+y^2+z^2+w^2=1 \} \setminus \{(x,y,0,0)\mid x^2+y^2 =1\}$ (i.e.$S^3 \setminus \text{planar unknot}$) by the homotopy $H(t,\vec{v}) = (1-t)\vec{v} + t \vec{v}/\|\vec{v}\|$
