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Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. $T_0$ (below) is called trivial. I conjecture that $NS(T)$ and $D_+(T)$ determine the triviality of $T$.

For that consider framed tangles only, since otherwise $NS(T), D_+(T)$ are trivial both for $T_0$ and for $T_2=$ the 2-braid with 2 negative crossings, and so $T_2$ can't be distinguished from $T_0.$

However, I conjecture that if $NS(T)$ is the framed unknot and $D_+(T)$ is the uknot with a positive kink, then $T$ is obtained from $T_0$, by adding framing $n$ to one strand and framing $-n$ to the other strand, for some $n$.

Would you have a suggestion for a proof?

Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. $T_0$ (below) is called trivial.

Question: Suppose that $NS(T)$ and $D_+(T)$ are trivial knots. Does it imply that $T$ is either $T_0$ or $T_2=$ the 2-braid with 2 negative crossings?

Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. I conjecture that if $NS(T)$ is the unknot and one of $D_+(T), D_-(T)$ is the uknot, then $T=T_0$.

We need to assume here that all tangles and links are framed (with the "blackboard" framing, i.e. the one parallel to your screen :-) since a double crossing $T$ would be a counterexample otherwise.

Is it known? Would you have a suggestion for a proof?

Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. $T_0$ (below) is called trivial. I conjecture that $NS(T)$ and $D_+(T)$ determine the triviality of $T$.

For that consider framed tangles only, since otherwise $NS(T), D_+(T)$ are trivial both for $T_0$ and for $T_2=$ the 2-braid with 2 negative crossings, and so $T_2$ can't be distinguished from $T_0.$

However, I conjecture that if $NS(T)$ is the framed unknot and $D_+(T)$ is the uknot with a positive kink, then $T$ is obtained from $T_0$, by adding framing $n$ to one strand and framing $-n$ to the other strand, for some $n$.

Would you have a suggestion for a proof?

Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. I conjecture that if $NS(T)$ is the unknot and one of $D_+(T), D_-(T)$ is the uknot, then $T=T_0$.

We need to assume here that all tangles and links are framed (with the "blackboard" framing, i.e. the one parallel to your screen :-) since a single crossing $T$ would be a counterexample otherwise.

Is it known? Would you have a suggestion for a proof?

Let $NS(T), D_+(T), D_-(T)$ denote closures of a $2$-tangle $T$ as in the picture. I conjecture that if $NS(T)$ is the unknot and one of $D_+(T), D_-(T)$ is the uknot, then $T=T_0$.

We need to assume here that all tangles and links are framed (with the "blackboard" framing, i.e. the one parallel to your screen :-) since a double crossing $T$ would be a counterexample otherwise.

Is it known? Would you have a suggestion for a proof?