We know that every link in $S^3$ is framed cobordant to the unknot with some framing. The idea is to study smooth homotopy classes of maps from $S^3$ to $S^2$. Actually in the title I have given $\mathbb R^3$ but any knot in $S^3$ can be isotoped to miss a point $p$ on $S^3$ and hence lie in $\mathbb R^3\cong S^3-p$
The framing given below is one where the frame twists once as it goes one round along the knot, i.e., framing given by $1\in\pi_1(SO(2))$
Does the disjoint union of $n$ (mutually unlinked) unknots with framing $1\in SO(2)$ represent the unknot with framing $n\in\pi_1(SO(2))$? Somehow I find it difficult to picture whether this is true or not. The only way I can perhaps argue is that in the framed cobordism classes of links form a group and the addition in the group is just the unlinked disjoint union. Is this true?
If false, is there any other way to represent the class of the unknot with framing with $n\in\pi_1(SO(2))$, by a link which has framing with one twist ($1\in SO(2)$) on each knot in the link?