Isotopy of links

Question

If a link $L$ has $\mu$ components $K_1,\ K_2,\ ... K_\mu$ and $L'$ has components $K'_1,\ K'_2,\ ... K'_\mu$ components, does "$L$ is isotopic to $L'$ " imply that "each $K_i$ is isotopic to $K'_{\sigma(i)}$ for some permutation $\sigma$"?

Answer 1

Essentially, yes. Formally, the isotopy has to be ambient. That is, $L$ and $L'$ are subsets of some manifold $M$; to say that $L$ is isotopic to $L'$ is to say that there is a homeomorphism $H$ of $M$ which is isotopic to the identity and which has $H(L) = L'$. In particular, this gives a knot-isotopy of each component of $L$ to some component of $L'$.

I've never seen the ordering of the components matter, but I'm not an expert in this field.