# Map into knot complement inducing quasi-isomorphism

I'm looking at the following topology qualifying exam question:

Let K be a knot in $S^3$. Construct a map $f: S^2 \vee S^1 \rightarrow (\mathbb{R}^3 - K)$ that induces an isomorphism of integral homology.

My biggest problem is I'm not really sure what a knot in $S^3$ is. Is it just an embedding of a circle into $S^3$? Is there a nice reference for these? Also, a hint as to what the map should be would be appreciated. Should I define it on a cellular level?

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Indeed, a knot is just a copy of $S^1$ in $S^3$. Most frequently, one considers only tame knots, which we can define variously to be the polygonal or smooth ones (of course, the two options are not equivalent, but they capture more or less the same class of examples), excluding the so called wild knots.
Try to construct such a map when $K$ is a standardly embedded circle (the so called unknot), then try a simple knot like the trefoil (google will find a picture for you), and by then you will know how to deal with the general case.
Thanks. I guess one trouble I'm having is visualizing $S^3$. I don't see why we're considering $R^3 - K$. Isn't $S^3$ in $R^4$? – Joe Dec 22 '11 at 20:54
Or is talking about $R^3$ as the boundary of $S^3$? – Joe Dec 22 '11 at 20:56
Visualize $S^3$ as $\mathbb R^3$ plus a point. A knot does not cover all of $S^3$, so you can always remove a point from $S^3-K$, and then you are left with $\mathbb R^3-K$, which you can draw without problems. – Mariano Suárez-Alvarez Dec 22 '11 at 20:57
@MarianoSuárez-Alvarez The statement is false is the ambient space is taken to be $S^3$ since $S^3\setminus K$ will have homology isomorphic to that of $S^1$. The second homology class is Alexander dual to the missing point at infinity. – Adam Dec 23 '11 at 1:35