I was trying to figurate what is the meaning of homotopy equivalence of a pair. In particular if we consider $X_1\subseteq X$ and $Y_1\subseteq Y$, and we have $(X,X_1)$ is homotopy equivalent to$(Y,Y_1)$, is this condition stronger then imposing only $X_1\simeq X$ and $Y_1 \simeq Y$?
I would like to know if there are some examples in which the pairs carry more structure then the single spaces. For example if we consider homeomorphism in place of homotopy equivalences, then $((\mathbb{R}^3,f(S^1))$ for different embeddings $f$ is not homeomorphic as a pair unless the two knot are equivalent. Does this example work also if we replace homotopy equivalent to homeomorphic?
Remark: I just want to add a little motivation to this question. I was interested in finding a relative version of the Whitehead theorem i.e. something like: if $(X,X_1), (Y,Y_1)$ are CW pairs and if there is a function that induces an isomorphism on every $\pi_n(X) \simeq \pi_n(Y),\;\pi_n(X_1) \simeq \pi_n(Y_1),\; \pi_n(X,X_1) \simeq \pi_n (Y,Y_1)$ then there is a homotopic equivalence between pairs $(X,X_1)\simeq (Y,Y_1)$. I strongly suspect this to be false because by the five lemma we can deduce the isomorphism between $\pi_n(X,X_1) \simeq \pi_n (Y,Y_1)$ from that of $\pi_n(X) \simeq \pi_n(Y),\;\pi_n(X_1) \simeq \pi_n(Y_1)$. The previous example would give a counterexample because each single space is homeomorphic, so the long exact sequences in homotopy are isomorphic but the two spaces are not homotopic equivalent as pairs.
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 $\pi_1(T)$.