The complement of an arc in $S^3$ is contractible

Question

Let $f:[0,1]\to S^3$ be an embedding, so that its image $X:=f([0,1])$ is an arc. How can we show that its complement $S^3-X$ is contractible? This is claimed in an answer of this question Homology groups of the complement of a knot in $S^3$ that I asked before.

Answer 1

By deforming the embedded path to a point, we can equivalently (up to homotopy) consider $S^3\setminus\{x\}$ for a point $x$. You can also consider $S^3$ to be the one-point compactification of $\Bbb R^3$, i.e. $\Bbb R^3\cup\{\infty\}$. The homotopy type of $S^3\setminus\{x\}$ is the same as $\left(\Bbb R^3\cup\{\infty\}\right)\setminus\{\infty\}$, which is homotopy equivalent to $\Bbb R^3$ and thus contractible.

If that sounds over-complicated, all I'm really saying is that $S^n$ minus a point is contractible for any $n$ (and any point). You can also see this from the cell structure of $S^n$ using one zero cell and one $n$-cell. If you delete an interior point of the $n$-cell, you can deform $D^n$ to $S^{n-1}$, which is glued via the trivial map to the zero cell.