Does homotopy equivalence of pairs distinguish knots?

Solution 1:

I don't know knot theory, although I suspect your suspect is correct. Below is an easier example that shows a pair indeed carries more information.

Let $T,T'$ be two torus, $S$ be a meridian circle in $T$ and $S'$ be a null-homotpic circle in $T'$; denote by $i,i'$ the corrsponding inclusion map. Suppose there are $f:(T,S)\rightarrow(T',S')$ and $g:(T',S')\rightarrow(T,S)$ such that $fg\simeq 1$ and $gf\simeq 1$. We abuse notation and use $f_*$ to mean either the induced map $\pi_1(S)\rightarrow\pi_1(S')$ or the map $\pi_1(T)\rightarrow\pi_1(T')$. In either case $f_*g_*$ and $g_*f_*$ are identities, so $f_*$ and $g_*$ are isomorphisms. If we denote by $\alpha$ a generator of $\pi_1(S)$, then $f_*(\alpha)$ is a generator of $\pi_1(S')$. However $i'_*$ is clearly trivial, so $g_*i'_*f_*(\alpha)$ is zero in $\pi_1(T)$. On the other hand $g_*i'_*f_*(\alpha)=i_*g_*f_*(\alpha)=i_*(\alpha)$, contradicting that $i_*(\alpha)$ is nonzero in $\pi_1(T)$.