Hyperbolization Theorem I know that for experts this question is trivial, but it's been a while I'm having trouble understanding this...
The version of the hyperbolization theorem I found on Aschenbrenner, Friedl and Wilton's book "3-manifold groups" is that a compact, orientable, irreducible 3-manifold with empty or torus boundary that is atoroidal and not homeomorphic to $\mathbb{S}^1\times \mathbb{D}$, $\mathbb{T}^2\times I$ or $\mathbb{K}^2\widetilde{\times} I$ and has infinite fundamental group is hyperbolic (in the sense that it also has finite volume). (To make it clear, $\mathbb{S}^1$ is the unit circle, $\mathbb{T}^2 = \mathbb{S}^1\times \mathbb{S}^1$ is the 2-torus, $\mathbb{K}^2$ is the Klein bottle, $I = [0,1]$ and $\mathbb{K}^2\widetilde{\times} I$ is the twisted, oriented $I$-bundle over $\mathbb{K}^2$, something I particularly do not understand well, which might be the reason I will ask my question...)
However, in many papers people prove hyperbolicity of a compact manifold with empty or torus boundary by showing there is no essential spheres, disks, tori and annuli. I know all these conditions are necessary for a manifold to be hyperbolic.
Here is my question: Why do we need to show there is no essential annuli in the manifold? Is it just to rule out the $\mathbb{K}^2\widetilde{\times} I$ case or am I missing something?
Thanks in advance and my apologies if the question is too trivial.
 A: I'm going to go through the essential surface hypotheses you give and show how they correspond to the hypotheses of the Hyperbolization Theorem.  (In short: essential annuli exclude the twisted $I$-bundle over the Klein bottle as well as the Cartesian product of a pair of pants with $S^1$, which is not atoroidal in the sense of the book.)
Essential spheres correspond directly to $M$ not being irreducible, so there is not much to say here.  Because you mention necessary and not just sufficient conditions, I will say this:  The Killing-Hopf theorem implies that (the interior of) a hyperbolic $3$-manifold has universal cover $\mathbb{H}^3$.  Since hyperbolic space is irreducible, then so are hyperbolic manifolds.
If there were an essential disk $D$, it would meet some torus boundary component along either a separating or nonseparating loop.  If nonseparating, the sphere boundary of $M-\nu(D)$ can be pushed in a bit to give $S^1\times D^2$ as a connect summand, and primality would imply $M=S^1\times D^2$, which is excluded by hypothesis.  Otherwise, if separating, it bounds a disk in the boundary, so that disk and $D$ form an $S^2$ that must bound a ball by irreducibility of $M$, hence there are no such disks.
Excluding essential tori is already part of the hypothesis of the Hyperbolization Theorem.  (However: atoroidal means something slightly different in that book, that any map $T^2\to M$ that is $\pi_1$-injective is homotopic to a map $T^2\to \partial M$.  That said, this is stronger than the definition that embedded incompressible tori are boundary-parallel.)
If there were an essential annulus $A$, then it either meets a pair of boundary components or a single boundary component. If it were a pair of boundary components, then there are the cases depending on whether each boundary of $A$ is separating or nonseparating in the respective boundary component.  If either is separating, then the corresponding disk, pushed into $M$ a bit, would compress $A$, contrary to $A$ being essential. If both are nonseparating, then by doing a similar pushoff of the boundary of $M-\nu(A)$, we get an embedded torus $T$.  If $T$ were compressible, a compressing disk would have to be on the side opposite the boundary components since the torus is $\pi_1$-injective on the boundary side.  Compressing along that disk gives an embedded sphere, which by irreducibility bounds a ball, hence $T$ bounds a solid torus.  With $X$ being a three-times punctured sphere (a pair of pants), $M$ can be identified as $X\times S^1$ with a solid torus glued in.  The way in which the meridian disk of the solid torus is glued in is in correspondence with a rational number: $a/b$ denotes the slope which is $b$ times around the $S^1$ direction and $a$ times around the $\partial X$ direction, so for example slope-$0$ is the disk glued in along an $S^1$ fiber and slope-$(1/0)$ is along a $\partial X$ component.  The slope-$0$ gluing is $S^3$ minus a split unlink, so it is not irreducible.  The slope-$(1/0)$-gluing is $T^2\times I$, which is an excluded case.  Gluing with any other slope is a Seifert-fibered space, and one can check that this space is not atoroidal in the book's sense (essentially for the same reason as why $X\times S^1$ is not atoroidal, see below).  On the other hand, if $T$ were incompressible, then it must either bound a solid torus (already handled) or be boundary-parallel.  Then $M$ can be identified as $X\times S^1$, which is not atoroidal in the book's sense, as will be explained below.
Now we consider if $\partial A$ is in a single boundary component $T$.  Again both curves of $\partial A$ are non-separating.  By using our knowledge of simple curves on the torus, we know they must be parallel.  Two cases: either the curves have the same induced orientation, or opposite.  In the case of opposite induced orientation, then by considering the two annuli in the boundary, we get two tori by combining each individually with $A$, which we then push in a bit from the boundary and away from each other so they do not intersect.  Neither bounds a solid torus since $A$ is essential, so both are boundary parallel.  Neither is parallel to $T$ since that would imply $A$ is not essential, hence $M$ is $X\times S^1$ for $X$ a closed twice-punctured disk (a pair of pants). While this has no essential tori, it is not atoroidal in the sense of the book: take an immersed loop in $X$ that goes around the points in opposite directions and cross it with $S^1$ to get a map $T^2\to X\times S^1$ that is $\pi_1$-injective but not homotopic to the boundary, since any homotopy would give such a homotopy for the loop in $X$.  (In the arXiv version of the book, they note that having three total boundary components and cone points is when the definitions diverge.  I will have to check whether this is a "small Seifert-fibered space," which it shouldn't be since it is Haken.  Edit: This has been clarified in this question.)
Lastly, $A$ meets the boundary component $T$ in curves of the same orientation, which is hard to imagine in totality, so here's an attempt at a picture:

This is a cross-section where the front and back are meant to be identified.  By taking $A$ with either annulus in $T-\partial A$ like before and pushing in, we get a Klein bottle $K$.  The regular neighborhood $\nu(K)$, since $K$ is non-orientable in an orientable manifold, is the twisted $I$ bundle over $K$, and $\partial\nu(K)$ is the oriented double cover, which is a torus.  The $\nu(K)$ side of $\partial\nu(K)$ is certainly not a solid torus (since $\pi_1(K)\neq\mathbb{Z}$), and the other side is not a solid torus because it contains $T$.  Since $M$ is atoroidal, $M=\nu(K)$, which is one of the excluded cases.
