fundamental class of the boundary of a knot complement

Question

Let $K$ be an oriented knot and let $F:\overline{B^2}\times K\rightarrow S^3$ be a thickening (with self linking number $0$; we do not explicitly need this as we work only with meridians of $K$, but lets take such a thickening for safety reasons). Denote by $X_K= S^3\setminus(B^2\times K)$ the knot complement. Then by my professors script we have $$F_\star([S^1\times K])=[\partial F(\overline{B^2}\times K)]=\color{red}{-}[\partial X_K].$$ Could someone explain the minus sign in the end to me? It seems wrong to me, as the fundamental class is determined by the given orientation on the manifold. For this we equip $S^3$ with an orientation by saying that a basis $v_1,\dots,v_n\in T_PS^3$ is positive iff $det(P,v_1,\dots,v_n)>0$. We equip all codimension $0$ submanifolds with that orientation, i.p. we equip the knotcomplement and thus also its boundary with the induced orientation. Since the boundary map maps the fundamental class onto the fundamental class of the boundary the above minus sign seems wrong to me or it should at least depend on the orientation of $K$ as I assume we want $F({0}\times K)$ to be an orientation preserving diffeomorphism.
Any help is greatly appreciated!

Answer 1

Something to remember about the induced orientation of the boundary $\partial M$ of an oriented manifold $M$ is that it's the one such that a basis $v_1,\dots,v_{n-1}\in T_p\partial M$ for $p\in \partial M$ is positive iff the basis $v_1,\dots,v_n,v_{\mathrm{out}}\in T_pM$ is positive, where $v_{\mathrm{out}}$ is any vector pointing "out" of the manifold. This means that if $N$ is a manifold that is decomposed into submanifolds $M_1$ and $M_2$ with $M_1\cap M_2=\partial M_1 = \partial M_2$, then $\partial M_1$ and $\partial M_2$ have opposite induced orientations. In particular, $[\partial M_1]=-[\partial M_2]$. An algebraic point of view is that the decomposition gives $[M_1] + [M_2] = [N]$, so, assuming $N$ has empty boundary, by taking boundaries we get at the level of chains $$[\partial M_1] + [\partial M_2] = \partial [M_1] + \partial[M_2] = \partial [N] = [\partial N] = 0.$$

We should assume that $F$ is an orientation-preserving map. Then, $\operatorname{im}(F)$ and $X_K$ constitute a decomposition of $S^3$ of the above type, so $[\partial \operatorname{im}(F)] = -[\partial X_k]$, which gives the identity you're asking about.

If you want to think about this in a more hands-on way, the image of each $S^1$ in $S^1\times K$ through $F$ is a meridian of $K$, so the image of $S^1\times K$ has an orientation where a positively oriented basis at a given point is given by $v_{\mu},v_{\lambda}$, with $v_{\mu}$ pointing with respect to the meridian orientation and $v_{\lambda}$ pointing with respect to the knot's orientation -- if you are holding onto the knot (envisioned as a thick rope) with your right hand, where your thumb is along the knot pointing in the direction of its orientation, then the rest of your fingers are wrapped around the knot in the meridian direction. The third vector for $S^3$ that completes the two vectors into a positive basis is one that is pointing out of the back of your hand, away from the knot. However, the induced orientation for $\partial X_K$ would have it that, to form a positive basis, this third vector should be pointing toward the knot instead.