Mapping class group of $S^1 \times S^1 \times I$ I am interested in the computation of the mapping class group of the manifold $M=S^1 \times S^1 \times I$. One can visualize $M$ as a "torus cross $I$", or as an "annulus cross $S^1$". $M$ is naturally Seifert fibered. I am considering here the group if isotopy classes of diffeomorphisms of $M$ which fix the boundary $\partial M = S^1 \times S^1 \amalg S^1 \times S^1$ pointwise. The latter can be formalized as $G = \pi_0Diff(M, \partial)$. 
There is a injection $\pi_1(Diff(T), Id) \rightarrow G$ where $T$ is the torus $T = S^1 \times S^1$. It is obtained by considering the homotopy class of a path $\lambda : I \rightarrow Diff(T)$ as the isotopy class of a diffeomorphism $\lambda : T \times I \rightarrow T \times I$ which happens to preserve each "layer of torus" of $M$ (more precisely, they preserve the fibers of the fibering over $I$). The diffeomorphism $\lambda$ one obtains this way fixes the boundary because the original path is a loop based at the identity. 
I suspect this morphism to be an isomorphism but this is merely an intuition. Thanks to this last consideration, the problem boils down to knowing if every diffeomorphism of $M$ fixing the boundary is isotopic to one that preserves the "torus layering" as well. This is why I mentioned the Seifert fibering of $M$, since such results exist for Seifert fiberings. 
Any insight is welcome ! Cheers. 
 A: Injectivity was not so obvious to me, so let me first provide a proof of this. Your map $
f:\pi_1(\mathrm{Diff}(T^2)) \to \pi_0(\mathrm{Diff}(T^2 \times I,\partial))$ has a left inverse $r:\pi_0(\mathrm{Diff}(T^2 \times I,\partial)) \to \pi_1(\mathrm{Diff}(T^2))$ defined as follows:
Let $\gamma(t)=((0,0),t) \in T^2 \times I$, and let $\pi:T^2 \times I \to T^2$ be the projection onto the torus factor. For any map $\phi \in \mathrm{Diff}(T^2 \times I,\partial)$, we have that $\pi(\phi\gamma(t))$ is a loop in $T^2$ based at $(0,0)$. In other words, we get a map $r:\pi_0(\mathrm{Diff}(T^2 \times I,\partial)) \to \pi_1(T^2)$. Since the identity component of $\mathrm{Diff}(T^2)$ is homotopy equivalent to $T^2$, we really have a map $r:\pi_0(\mathrm{Diff}(T^2 \times I,\partial)) \to \pi_1(\mathrm{Diff}(T^2))$. It's not hard to see that $rf=\mathrm{id}$, which shows that $f$ is injective.
As for surjectivity, take some $\phi \in \mathrm{Diff}(T^2 \times I,\partial)$. By postcomposing with some map in the image of $f$, we may assume that $\phi$ fixes the fiber $(0,0) \times I$ up to isotopy. Now $\phi$ also gives a $\mathbb{Z}^2$-equivariant map $\tilde \phi: \mathbb{R}^2 \times I \to \mathbb{R}^2 \times I$ on the universal cover, and we must have that this map fixes both components of the boundary. Using the straight-line homotopy, we have $\tilde \phi \simeq \mathrm{id}\ \mathrm{rel}\ \partial$ in a $\mathbb{Z}^2$ equivariant way. This descends to a homotopy $\phi \simeq \mathrm{id}\ \mathrm{rel}\ \partial$ on $T^2 \times I$. In general, this is not enough to say that $\phi$ is isotopic to the identity. But in the the case of Haken $3$-manifolds  (such as $T^2 \times I$), it is.
Remark: this is basically exactly the same as the proof that the mapping class group of the annulus $\pi_0(\mathrm{Diff}(S^1 \times I,\partial))$ is isomorphic to $\mathbb{Z} = \pi_1(\mathrm{Diff}(S^1))$.
