Do homeomorphisms of boundary components of 3-manifolds extend to the manifold?

Question

The question that I'd like to answer can be generalized to the following: if $M$ is an orientable 3-manifold and $F$ is a boundary component of $M$ (which may have other boundary components), can an arbitrary diffeomorphism of $F$ with itself be extended to a diffeomorphism of $M$ with itself? I would be very surprised if the answer to the question in this generality is yes, so I'll ask a more specific question.

If $M$ is a complement of a link with 1 or 2 components in $S^3$ and $\sigma:T^2 \to T^2$ is the antipodal map in each component, is there a diffeomorphism $\sigma':M \to M$ that restricts to $\sigma$?

I don't really know how to approach this, but here's a couple thoughts:

1) If I understand right, this question can be reduced to the question of whether the subgroup of the mapping class group of $M$ that preserves $F$ surjects onto the mapping class group of $F$ (since it's my impression that isotopies can be extended from submanifolds).

2) If we lower the dimension by 1 it seems like the answer is yes, at least if the last comment was right. To see this, let $C$ be the boundary component that is to be sent to itself, and write the surface as a quotient of a polygon in $\mathbb R^2$ that is symmetric about the origin. Also, center the boundary component $C$ around the origin, and put the other boundary components symmetrically around the origin. Then the antipodal map in $\mathbb R^2$ induces a diffeomorphism of $F$ with the desired properties.

(I'm not sure about tags - retag as appropriate.)

Answer 1

The more general question has a negative answer for the following reason (which could probably be extended to the special case with more thought). The mapping class group of a finite volume hyperbolic 3-manifold is finite by Mostow rigidity. But finite volume hyperbolic 3-manifolds can have tori as boundary components, whose mapping class group is infinite. So in this case most diffeomorphisms of a boundary component cannot be lifted to the 3-manifold.

Answer 2

I interprete the "antipodal map" $T^2 \to T^2$ as the elliptic involution, which acts as multiplication by $-1$ on homology. For many link complements the elliptic involution extends to the link complement, but not for all. This holds if and only if the link is strongly invertible: the Wikipedia page furnishes some counterexamples.

The elliptic involution in fact extends to the solid torus neighborhood of each knot component of the link via a map that inverts the orientation of the knot, and therefore you get a map $S^3 \to S^3$ which preserves the link but inverts the orientation of every knot component.